> >>
40yahoogroups.com>,
> >> "Louis P."
> >> > > <rockprog80@> wrote:
> >> > > >
> >> > > > Hi,
> >> > > >
> >> > > > @Tony:Sorry I thought gradient was the good word. Maybe I meant
> >> > > angle, or
> >> > > > degree? I want a LR slope of approx. 30-50 degrees. How can
I do
> >> that?
> >> > > > @gp_sydney: I need to go to work, but will check this
tomorrow or
> >> > > Monday. I
> >> > > > looked very quickly and was wondering where in that formula
can I
> >> > > determine
> >> > > > how steep the slope will be. That would help a lot!
> >> > > >
> >> > > > Thank you you two a lot!
> >> > > >
> >> > > > Louis
> >> > > >
> >> > > >
> >> > > >
> >> > > > 2008/9/20 gp_sydney <gp.investment@>
> >> > > >
> >> > > > > Sorry, a couple of typos in the formula. It should be:
> >> > > > >
> >> > > > >
> >> > > > > n = HHVBars(High, periods);
> >> > > > > avX = MA(x, n);
> >> > > > > avY = MA(y, n);
> >> > > > > avXY = MA(x*y, n);
> >> > > > > avXX = MA(x*x, n);
> >> > > > >
> >> > > > > b = (avXY - avX*avY) / (avXX - avX*avX);
> >> > > > > a = avY - b*avX;
> >> > > > >
> >> > > > > GP
> >> > > > >
> >> > > > >
> >> > > > > --- In
amibroker@xxxxxxxxxxxxxxx
<amibroker%
40yahoogroups.com>
> >> <amibroker%
40yahoogroups.com><amibroker%
> >>
> >> > >
40yahoogroups.com>,
> >> > >
> >> > > > > "gp_sydney" <gp.investment@> wrote:
> >> > > > > >
> >> > > > > > Louis,
> >> > > > > >
> >> > > > > > I gather you want to use regression to get a best-fit line
> >> from the
> >> > > > > > most-recent HHV, not just draw a line between the end
points.
> >> > > From my
> >> > > > > > understanding of regression though, which isn't a lot, I
> >> gather that
> >> > > > > > such a line may not actually pass through either end
point, so
> >> > > may not
> >> > > > > > go "from the HHV" exactly or end at the current bar exactly
> >> (to get
> >> > > > > > that, just draw a line between those two points as
previously
> >> > > > > > mentioned). Once you have the line though, you could
offset it
> >> > > > > > vertically to get it to pass exactly through the HHV or
> >> current bar
> >> > > > > > (but not both of course).
> >> > > > > >
> >> > > > > > From a response Tomasz gave you in another thread about
how to
> >> > > > > > calculate Rsquared without a loop:
> >> > > > > >
> >> > > > > >
http://finance.groups.yahoo.com/group/amibroker/message/129392
> >> > > > > >
> >> > > > > > you can similarly do the regression itself. According
to Tomasz,
> >> > > > > > LinRegSlope already accepts variable periods (ie. arrays),
> >> so to get
> >> > > > > > the slope you should be able to just use that function.
Or to
> >> > > > > > calculate both coefficients yourself, then something
like this
> >> > > (based
> >> > > > > > on the formula given at
http://www.educalc.net/2104083.page):
> >> > > > > >
> >> > > > > > n = HHVBars(High, periods);
> >> > > > > > avX = MA(x, n);
> >> > > > > > avY = MA(y, n);
> >> > > > > > avXY = MA(x*y, n);
> >> > > > > > axXX = MA(x*x, n);
> >> > > > > >
> >> > > > > > b = (avXY - avX*avY) / (avXX - avX*avX);
> >> > > > > > a = axY - b*avX;
> >> > > > > >
> >> > > > > > Hopefully I've got that right (I haven't tested it).
The line
> >> > > formula
> >> > > > > > is then:
> >> > > > > >
> >> > > > > > line = a + b*x;
> >> > > > > >
> >> > > > > > The slope is 'b' and the intercept 'a'. The slope is in
> >> dollars per
> >> > > > > > bar. I'm not sure what you want by a percentage slope
> >> though, as the
> >> > > > > > 'x' and 'y' axes are in completely different units.
> >> > > > > >
> >> > > > > > Also note that the calculated lines would only be
straight on a
> >> > > linear
> >> > > > > > price scale (if they were plotted). If you use semi-log,
> >> then the
> >> > > > > > formula for a straight line is different - and how to do
> >> regression
> >> > > > > > for it would take some figuring out.
> >> > > > > >
> >> > > > > > Hope this helps.
> >> > > > > >
> >> > > > > > Regards,
> >> > > > > > GP
> >> > > > > >
> >> > > > > >
> >> > > > > > --- In
amibroker@xxxxxxxxxxxxxxx
<amibroker%
40yahoogroups.com>
> >> <amibroker%
40yahoogroups.com><amibroker%
> >>
> >> > >
40yahoogroups.com>,
> >> > >
> >> > > "Louis
> >> > > > > P." <rockprog80@> wrote:
> >> > > > > > >
> >> > > > > > > Hi,
> >> > > > > > >
> >> > > > > > > I need the gradient of the slope, and for each bar. This
> >> is where
> >> > > > > it is
> >> > > > > > > difficult...
> >> > > > > > >
> >> > > > > > > Thanks,
> >> > > > > > >
> >> > > > > > > Louis
> >> > > > > > >
> >> > > > > > > 2008/9/19 Tony Grimes <Tonez.Email@>
> >> > > > > > >
> >> > > > > > > > Louis,
> >> > > > > > > >
> >> > > > > > > > If your looking for the slope & difference between
HHV of 20
> >> > > bars
> >> > > > > > & the
> >> > > > > > > > current close, all you should need is this:
> >> > > > > > > >
> >> > > > > > > > Pds=20;
> >> > > > > > > >
> >> > > > > > > > LastHighBar = HHVBars(High, Pds);
> >> > > > > > > > LastHighVal = HHV(High, Pds);
> >> > > > > > > >
> >> > > > > > > > Slope = IIf(LastHighBar,(Close - LastHighVal) /
> >> LastHighBar,0);
> >> > > > > > > > CloseDiff = Close - Ref(Close, -LastHighBar);
> >> > > > > > > >
> >> > > > > > > >
> >> > > > > > > > On Fri, Sep 19, 2008 at 4:51 PM, Louis P. <rockprog80@>
> >> wrote:
> >> > > > > > > >
> >> > > > > > > >> Hi Tony,
> >> > > > > > > >>
> >> > > > > > > >> Thank you a lot for your response. I'm still very
weak with
> >> > > > > > loops. Last
> >> > > > > > > >> time experimented with one, I had to reboot my
> >> computer! :) So
> >> > > > > > do you know
> >> > > > > > > >> how such a loop could work? And if I run, let's say 2
> >> minutes
> >> > > > > > bar for one
> >> > > > > > > >> year, wouldn't that be really really long to deal
with?
> >> I have
> >> > > > > > PIV with 1.5
> >> > > > > > > >> GHz ram.
> >> > > > > > > >>
> >> > > > > > > >> I am looking for a line that ends at each bar and that
> >> starts
> >> > > > > > from the HHV
> >> > > > > > > >> of 20 bars, and I want to do things with this bar:
e.g.
> >> compare
> >> > > > > > the closes
> >> > > > > > > >> between current bar and the HHV to the bar and
> >> establish the
> >> > > > > > gradient of
> >> > > > > > > >> that linear regression bar for each bar.
> >> > > > > > > >>
> >> > > > > > > >> Thanks a lot!
> >> > > > > > > >>
> >> > > > > > > >> Louis
> >> > > > > > > >>
> >> > > > > > > >> 2008/9/19 Tony Grimes <Tonez.Email@>
> >> > > > > > > >>
> >> > > > > > > >>> Hi Louis,
> >> > > > > > > >>>
> >> > > > > > > >>> A loop will work, but how slow - it depends
(Speed of your
> >> > > > > computer,
> >> > > > > > > >>> number of bars loaded, how many loops your using
etc...).
> >> > > > > > Without seeing
> >> > > > > > > >>> what your actually looking for (The end result), I
> >> think you
> >> > > > > > could do it
> >> > > > > > > >>> with one loop, with only one pass through the loop.
> >> The speed
> >> > > > > > should be OK.
> >> > > > > > > >>>
> >> > > > > > > >>> Good Luck.
> >> > > > > > > >>>
> >> > > > > > > >>> Tony
> >> > > > > > > >>>
> >> > > > > > > >>> On Fri, Sep 19, 2008 at 2:47 PM, Louis P.
> >> <rockprog80@> wrote:
> >> > > > > > > >>>
> >> > > > > > > >>>> Hi Tony,
> >> > > > > > > >>>>
> >> > > > > > > >>>> Thanks for the tips. Basically, I'd need a loop and
> >> use it on
> >> > > > > > each and
> >> > > > > > > >>>> every bar of the array to determine the LR, right?
> >> > > > > > > >>>>
> >> > > > > > > >>>> That will slow down my computer a lot, don't you
think?
> >> > > > > > > >>>>
> >> > > > > > > >>>> Thanks,
> >> > > > > > > >>>>
> >> > > > > > > >>>> Louis
> >> > > > > > > >>>>
> >> > > > > > > >>>> 2008/9/16 Tony Grimes <Tonez.Email@>
> >> > > > > > > >>>>
> >> > > > > > > >>>>> SelectedValue takes an array ( of numbers) and
returns a
> >> > > > > single
> >> > > > > > > >>>>> number based on the bar that is selected in the
chart.
> >> > > > > > > >>>>>
> >> > > > > > > >>>>> The first formula worked because SelectedValue was
> >> > > giving you a
> >> > > > > > > >>>>> number.
> >> > > > > > > >>>>>
> >> > > > > > > >>>>> Look at it this way: Array --> SelectedValue --->
> >> Number.
> >> > > > > > > >>>>>
> >> > > > > > > >>>>> Remove SelectedValue: Array---->Array.
> >> > > > > > > >>>>>
> >> > > > > > > >>>>> You can draw a line with single numbers, but
not arrays.
> >> > > > > > > >>>>>
> >> > > > > > > >>>>> You can always use a loop.
> >> > > > > > > >>>>>
> >> > > > > > > >>>>> You might want to read: Understanding how AFL
works,
> >> in the
> >> > > > > > Amibroker
> >> > > > > > > >>>>> users guide. Until you really understand AFL &
array
> >> > > > > > processing, you are
> >> > > > > > > >>>>> going to keep running into these problems, which
> >> will just
> >> > > > > > slow you down.
> >> > > > > > > >>>>>
> >> > > > > > > >>>>>
> >> > > > > > > >>>>> On Tue, Sep 16, 2008 at 10:34 PM, Louis P.
> >> > > <rockprog80@>wrote:
> >> > > > > > > >>>>>
> >> > > > > > > >>>>>> Hi Tony,
> >> > > > > > > >>>>>>
> >> > > > > > > >>>>>> Why was the first formula working (the one with
> >> > > > > > selectedvalue) and not
> >> > > > > > > >>>>>> the second one? Why simply deleting the
"selectedvalue"
> >> > > > > > makes it an array
> >> > > > > > > >>>>>> that will not be accept in "linearray"?
> >> > > > > > > >>>>>>
> >> > > > > > > >>>>>> Is there any way to draw a line without using
> >> lastvalue or
> >> > > > > > > >>>>>> selectedvalue? Do I need to use a loop?
> >> > > > > > > >>>>>>
> >> > > > > > > >>>>>> Thanks,
> >> > > > > > > >>>>>>
> >> > > > > > > >>>>>> Louis
> >> > > > > > > >>>>>>
> >> > > > > > > >>>>>> 2008/9/16 Tony Grimes <Tonez.Email@>
> >> > > > > > > >>>>>>
> >> > > > > > > >>>>>>> Louis,
> >> > > > > > > >>>>>>>
> >> > > > > > > >>>>>>> All of the variables you are creating for the
> >> LineArray
> >> > > > > > function are
> >> > > > > > > >>>>>>> arrays themselves. Although LineArray
generates an
> >> > > array, it
> >> > > > > > does not accept
> >> > > > > > > >>>>>>> any arrays as inputs. Additionally, your error
> >> message was
> >> > > > > > probably
> >> > > > > > > >>>>>>> different. It probably went from complaining
about
> >> > > argument
> >> > > > > > #4 having the
> >> > > > > > > >>>>>>> incorrect type (which you corrected) to argument
> >> #3 having
> >> > > > > > the incorrect
> >> > > > > > > >>>>>>> type.
> >> > > > > > > >>>>>>>
> >> > > > > > > >>>>>>> On Tue, Sep 16, 2008 at 9:10 PM, Louis P.
> >> > > <rockprog80@>wrote:
> >> > > > > > > >>>>>>>
> >> > > > > > > >>>>>>>> Hi,
> >> > > > > > > >>>>>>>>
> >> > > > > > > >>>>>>>> Thank you for your help.
> >> > > > > > > >>>>>>>>
> >> > > > > > > >>>>>>>> @Ara:
> >> > > > > > > >>>>>>>>
> >> > > > > > > >>>>>>>> If in
> >> > > > > > > >>>>>>>>
> >> > > > > > > >>>>>>>> barhh1 = HHVBars( High, Periods ) ;
> >> > > > > > > >>>>>>>> bi1 = BarIndex();
> >> > > > > > > >>>>>>>> y11 = LinearReg( C, barhh1 ) ;
> >> > > > > > > >>>>>>>> y01 = LinRegIntercept( C, barhh1 ) ;
> >> > > > > > > >>>>>>>> sl1 = LineArray( bi1-barhh1+0, y01, bi1, y11, 0,
> >> True );
> >> > > > > > > >>>>>>>>
> >> > > > > > > >>>>>>>> I replace
> >> > > > > > > >>>>>>>>
> >> > > > > > > >>>>>>>> sl1 = LineArray( bi1-barhh1+0, y01, bi1, y11, 0,
> >> True );
> >> > > > > > > >>>>>>>>
> >> > > > > > > >>>>>>>> by
> >> > > > > > > >>>>>>>>
> >> > > > > > > >>>>>>>> sl1 = LineArray( bi1-barhh1+0, y01, bi1,
> >> > > LastValue(y11), 0,
> >> > > > > > True
> >> > > > > > > >>>>>>>> );
> >> > > > > > > >>>>>>>>
> >> > > > > > > >>>>>>>> I still have the same error message. I don't
know
> >> from
> >> > > > > > where it is
> >> > > > > > > >>>>>>>> coming.. unfortunately!
> >> > > > > > > >>>>>>>>
> >> > > > > > > >>>>>>>>
> >> > > > > > > >>>>>>>> @gp_sydney:
> >> > > > > > > >>>>>>>>
> >> > > > > > > >>>>>>>> That was a typo, you are right; I arranged
that by
> >> > > adding a
> >> > > > > > 1. But
> >> > > > > > > >>>>>>>> still, the problem remains: the last line does
> >> not work.
> >> > > > > > > >>>>>>>>
> >> > > > > > > >>>>>>>> One day, I asked support if I needed a loop
to do
> >> such LR
> >> > > > > > and they
> >> > > > > > > >>>>>>>> said I should not need one.
> >> > > > > > > >>>>>>>>
> >> > > > > > > >>>>>>>> Here is the original code:
> >> > > > > > > >>>>>>>>
> >> > > > > > > >>>>>>>> barhh = SelectedValue( HHVBars( High,
Periods ) );
> >> > > > > > > >>>>>>>> bi = SelectedValue( BarIndex() );
> >> > > > > > > >>>>>>>> y1 = SelectedValue( LinearReg( C, barhh ) );
> >> > > > > > > >>>>>>>> y0 = SelectedValue( LinRegIntercept( C,
barhh ) );
> >> > > > > > > >>>>>>>> sl = LineArray( bi-barhh+0, y0, bi, y1, 0,
True );
> >> > > > > > > >>>>>>>>
> >> > > > > > > >>>>>>>> What I want to do is simply eliminate the
> >> "selectedvalue"
> >> > > > > > part and
> >> > > > > > > >>>>>>>> use the code not only for the selected data but
> >> for the
> >> > > > > > whole data. I want
> >> > > > > > > >>>>>>>> to be able to draw a line from each HHV to each
> >> bar and
> >> > > > > > then work with the
> >> > > > > > > >>>>>>>> result.
> >> > > > > > > >>>>>>>>
> >> > > > > > > >>>>>>>> If it can't be done without a loop, I feel like
> >> I'll be
> >> > > > > > lost in time
> >> > > > > > > >>>>>>>> again; last time I tried to run a loop on my
> >> computer it
> >> > > > > > freezed and after 2
> >> > > > > > > >>>>>>>> minutes I decided to shut down AB...
> >> > > > > > > >>>>>>>>
> >> > > > > > > >>>>>>>>
> >> > > > > > > >>>>>>>> Thanks for the help,
> >> > > > > > > >>>>>>>>
> >> > > > > > > >>>>>>>> Louis
> >> > > > > > > >>>>>>>>
> >> > > > > > > >>>>>>>>
> >> > > > > > > >>>>>>>>
> >> > > > > > > >>>>>>>>
> >> > > > > > > >>>>>>>>
> >> > > > > > > >>>>>>>>
> >> > > > > > > >>>>>>>>
> >> > > > > > > >>>>>>>> 2008/9/16 gp_sydney <gp.investment@>
> >> > > > > > > >>>>>>>>
> >> > > > > > > >>>>>>>>> As Ara said, in the shown code snippet you
don't
> >> have
> >> > > > > > "barhh"
> >> > > > > > > >>>>>>>>> defined,
> >> > > > > > > >>>>>>>>> only "barhh1".
> >> > > > > > > >>>>>>>>>
> >> > > > > > > >>>>>>>>> Beyond that, you have the same issue I
mentioned
> >> > > > > > originally, that
> >> > > > > > > >>>>>>>>> the
> >> > > > > > > >>>>>>>>> linear regression functions and LineArray
> >> function take
> >> > > > > scalar
> >> > > > > > > >>>>>>>>> values
> >> > > > > > > >>>>>>>>> (ie. single numbers) as parameters, not arrays.
> >> > > > > > > >>>>>>>>>
> >> > > > > > > >>>>>>>>> I gather you're trying to create a line
from the
> >> > > > > > most-recent HHV
> >> > > > > > > >>>>>>>>> value
> >> > > > > > > >>>>>>>>> using the subsequent close data for every
bar on the
> >> > > > > > chart. As I
> >> > > > > > > >>>>>>>>> don't
> >> > > > > > > >>>>>>>>> think the linear regression functions can take
> >> arrays
> >> > > > > for the
> >> > > > > > > >>>>>>>>> period,
> >> > > > > > > >>>>>>>>> I think you'd need to use a loop and do the
linear
> >> > > > > regression
> >> > > > > > > >>>>>>>>> yourself
> >> > > > > > > >>>>>>>>> at each bar (you could call the array functions
> >> > > within the
> >> > > > > > loop,
> >> > > > > > > >>>>>>>>> but
> >> > > > > > > >>>>>>>>> since they fill a whole array each time, they
> >> would do a
> >> > > > > > lot of
> >> > > > > > > >>>>>>>>> unnecessary work). If you do that yourself
> >> inside the
> >> > > > > > loop, then at
> >> > > > > > > >>>>>>>>> each bar you'd have scalar 'x' and 'y'
values to
> >> > > calculate
> >> > > > > > the line
> >> > > > > > > >>>>>>>>> slope and so on.
> >> > > > > > > >>>>>>>>>
> >> > > > > > > >>>>>>>>> For what it's worth, the BarIndex function
> >> simply gives
> >> > > > > > you the bar
> >> > > > > > > >>>>>>>>> number. It provides a way of using the
current bar
> >> > > number
> >> > > > > > in array
> >> > > > > > > >>>>>>>>> formula.
> >> > > > > > > >>>>>>>>>
> >> > > > > > > >>>>>>>>> Regards,
> >> > > > > > > >>>>>>>>> GP
> >> > > > > > > >>>>>>>>>
> >> > > > > > > >>>>>>>>>
> >> > > > > > > >>>>>>>>> --- In
> >> > >
amibroker@xxxxxxxxxxxxxxx <amibroker%
40yahoogroups.com>
<amibroker%
> >>
40yahoogroups.com><amibroker%
> >> > >
40yahoogroups.com>
> >> > > > > > <amibroker%
40yahoogroups.com>,
> >> > > > > > > >>>>>>>>> "Louis P." <rockprog80@> wrote:
> >> > > > > > > >>>>>>>>> >
> >> > > > > > > >>>>>>>>> > Hi,
> >> > > > > > > >>>>>>>>> >
> >> > > > > > > >>>>>>>>> > Sorry, You can replace "periods" by 50 if you
> >> wish. I
> >> > > > > > just forgot
> >> > > > > > > >>>>>>>>> to
> >> > > > > > > >>>>>>>>> > include that.
> >> > > > > > > >>>>>>>>> >
> >> > > > > > > >>>>>>>>> > barhh1 = HHVBars( High, *50* ) ;
> >> > > > > > > >>>>>>>>> > bi1 = BarIndex() ;
> >> > > > > > > >>>>>>>>> > y11 = LinearReg( C, barhh ) ;
> >> > > > > > > >>>>>>>>> > y01 = LinRegIntercept( C, barhh ) ;
> >> > > > > > > >>>>>>>>> > sl1 = LineArray( bi1-barhh1+0, y01, bi1,
y11, 0,
> >> > > True );
> >> > > > > > > >>>>>>>>> >
> >> > > > > > > >>>>>>>>> > Still, it is not working, even if barhh1 is
> >> defined...
> >> > > > > > > >>>>>>>>> >
> >> > > > > > > >>>>>>>>> > Louis
> >> > > > > > > >>>>>>>>> >
> >> > > > > > > >>>>>>>>> > 2008/9/16 Ara Kaloustian <ara1@>
> >> > > > > > > >>>>>>>>> >
> >> > > > > > > >>>>>>>>> > > y11 and y01 use "barhh" which is not
defined.
> >> > > > > > > >>>>>>>>> > >
> >> > > > > > > >>>>>>>>> > > You have defined "barhh1"
> >> > > > > > > >>>>>>>>> > >
> >> > > > > > > >>>>>>>>> > >
> >> > > > > > > >>>>>>>>> > >
> >> > > > > > > >>>>>>>>> > > ----- Original Message -----
> >> > > > > > > >>>>>>>>> > > *From:* Louis P. <rockprog80@>
> >> > > > > > > >>>>>>>>> > > *To:*
> >> > >
amibroker@xxxxxxxxxxxxxxx <amibroker%
40yahoogroups.com>
<amibroker%
> >>
40yahoogroups.com><amibroker%
> >> > >
40yahoogroups.com>
> >> > > > > > <amibroker%
40yahoogroups.com>
> >> > > > > > > >>>>>>>>> > > *Sent:* Tuesday, September 16, 2008 2:46 PM
> >> > > > > > > >>>>>>>>> > > *Subject:* [amibroker] What is wrong?
> >> > > > > > > >>>>>>>>> > >
> >> > > > > > > >>>>>>>>> > > Hi,
> >> > > > > > > >>>>>>>>> > >
> >> > > > > > > >>>>>>>>> > > What is wrong in the following formula?
> >> > > > > > > >>>>>>>>> > >
> >> > > > > > > >>>>>>>>> > > barhh1 = HHVBars( High, Periods ) ;
> >> > > > > > > >>>>>>>>> > > bi1 = BarIndex() ;
> >> > > > > > > >>>>>>>>> > > y11 = LinearReg( C, barhh ) ;
> >> > > > > > > >>>>>>>>> > > y01 = LinRegIntercept( C, barhh ) ;
> >> > > > > > > >>>>>>>>> > > sl1 = LineArray( bi1-barhh1+0, y01,
bi1, y11, 0,
> >> > > True );
> >> > > > > > > >>>>>>>>> > >
> >> > > > > > > >>>>>>>>> > >
> >> > > > > > > >>>>>>>>> > > Thanks,
> >> > > > > > > >>>>>>>>> > >
> >> > > > > > > >>>>>>>>> > > Louis
> >> > > > > > > >>>>>>>>> > >
> >> > > > > > > >>>>>>>>> > > p.s. There was "Selectedvalue" in the
first four
> >> > > lines
> >> > > > > > but I
> >> > > > > > > >>>>>>>>> don't
> >> > > > > > > >>>>>>>>> want to
> >> > > > > > > >>>>>>>>> > > plot it on the chart based on where I
am on that
> >> > > > > > chart, but
> >> > > > > > > >>>>>>>>> simply
> >> > > > > > > >>>>>>>>> set the
> >> > > > > > > >>>>>>>>> > > variable so I can use the stuff later.
> >> > > > > > > >>>>>>>>> > >
> >> > > > > > > >>>>>>>>> > >
> >> > > > > > > >>>>>>>>> > >
> >> > > > > > > >>>>>>>>> >
> >> > > > > > > >>>>>>>>>
> >> > > > > > > >>>>>>>>>
> >> > > > > > > >>>>>>>>
> >> > > > > > > >>>>>>>
> >> > > > > > > >>>>>>
> >> > > > > > > >>>>>
> >> > > > > > > >>>>
> >> > > > > > > >>>
> >> > > > > > > >>
> >> > > > > > > >
> >> > > > > > > >
> >> > > > > > >
> >> > > > > >
> >> > > > >
> >> > > > >
> >> > > > >
> >> > > >
> >> > >
> >> > >
> >> > >
> >> >
> >>
> >>
> >>
> >
> >
>