Position X Sort-> Please urgent!

Hi all,

I'm with problems with my transformation and I need to give my project in
next week, and need to make others things.
Well...
I've 2 xml documents and 1 xsl document.
The first doc xml (XMLORIGINAL.XML)
The second xml document (XMLADD.XML) have informations about first document.
The relashion this elements is for position, but I need to print this
context using the sort, and I'm having problems with this.
Exists a formal of to get a position of element in original formate?
My example:

<!-- XMLORIGINAL.XML -->
<?xml-stylesheet type="text/xsl" href="XSLORIGINAL.XSL"?>
<ORIGINAL>
  <TEST>1</TEST>
  <TEST>2</TEST>
  <TEST>3</TEST>
  <TEST>4</TEST>
  <TEST>5</TEST>
</ORIGINAL>


<!-- XMLADD.XML -->
<ADDS>
  <ADD>A</ADD> <!-- This element if referent the 1º <TEST> element -->
  <ADD>B</ADD> <!-- This element if referent the 2º <TEST> element -->
  <ADD>C</ADD> <!-- This element if referent the 3º <TEST> element -->
  <ADD>D</ADD> <!-- This element if referent the 4º <TEST> element -->
  <ADD>E</ADD> <!-- This element if referent the 5º <TEST> element -->
</ADDS>

<!-- XSLORIGINAL.XSL -->
<xsl:template match="ORIGINAL">
   <xsl:for-each select="TEST">
      <xsl:sort select="." data-type="number" order="descending"/>
          <xsl:value-of select="."/>
          <xsl:variable name="var_pos" select="position()"/>
          <xsl:value-of select="document('XMLADD.XML')/ADDS/ADD[$var_pos]">
   </xsl:for-each>
</xsl:template>


I need to print so:
------------------
5E
4D
3C
2B
1A

And my result is:
------------------
5A
4B
3C
2D
1E

If I can't to print so, then exist another form????


Thanks all,


Paulo.


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