Counting level of nodes beneath current (in XPATH)

Hi all,

I have an XML with the following rules (simplified):

X -> Y+
Y -> Z X?
Z -> (textnode)


An example is like this:
<X>
 <Y>
   <Z>lv 1</Z>
   <X>
     <Y>
       <Z>lv 2</Z>
     </Y>
   </X>
 </Y>
 <Y>
   <Z>lv 1</Z>
   <X>
     <Y>
       <Z>lv 2</Z>
       <X>
         <Y>
           <Z>lv 3</Z>
         </Y>
       </X>
     </Y>
     <Y>
       <Z>lv 2</Z>
       <X>
         <Y>
           <Z>lv 3</Z>
         </Y>
       </X>
       <X>
         <Y>
           <Z>lv 3</Z>
           <X>
             <Y>
               <Z>lv 4</Z>
             </Y>
             <Y>
               <Z>lv 4</Z>
             </Y>
           </X>
         </Y>
       </X>
     </Y>
   </X>
 </Y>
</X>


What I am interested in, is the level of Y nodes.
If I wanted the top two levels, I could easily do this in xpath using 
count(ancestor::Y).
The problem is that I cannot readily count all descendants - 
count(descendant::Y) will not count the number of levels but the total 
number of Y nodes that are beneath the current Y node.
Is there a way for such counting in XPath?

I want all except the buttom 1, 2, 3 or more levels. The exact amount 
determined from a parameter.
I could have something like [not(X)] for leaf nodes, [not(X/Y/X)] for the 
the buttom 2 levels (leaf nodes
and one level up, counting from the leaf). But as the number of levels are 
to be determined dynamically,
this solution is not the best:

<xsl:template match="/">
 <xsl:choose>
   <xsl:when test="$lv='1'">
     <xsl:apply-templates select="Y[not(X)]" mode="One"/>
   </xsl:when>
   <xsl:when test="$lv='2'">
     <xsl:apply-templates select="Y[not(X/Y/X)]" mode="Two"/>
   </xsl:when>
   ...
   <xsl:otherwise>
     <xsl:apply-templates select="Y" mode="Zero"/>
   </xsl:otherwise>
 </xsl:choose>
</xsl:template>

<xsl:template match="Y" mode="One">
 <!-- Do some stuff -->
 <xsl:apply-templates select="Y[not(X)]" mode="One"/>
</xsl:template>


Thank you!
Ragulf Pickaxe :-|

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