Hi I am using a breadcrumb on mysite generated by xslt and it populates the breadcrumb properly. I want to restrict the breadcrumb to 3 or 4 level of top nodes. Like if I am on 5th level of page and breadcrumb shows all the 5 pages, i just want to display the top 3. how can i do that in following xslt?

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<?xml version="1.0" encoding="UTF-8"?>
<!DOCTYPE xsl:stylesheet [ <!ENTITY nbsp "&#x00A0;"> ]>
<xsl:stylesheet
    version="1.0"
    xmlns:xsl="http://www.w3.org/1999/XSL/Transform"
    xmlns:msxml="urn:schemas-microsoft-com:xslt"
    xmlns:umbraco.library="urn:umbraco.library" xmlns:Exslt.ExsltCommon="urn:Exslt.ExsltCommon" xmlns:Exslt.ExsltDatesAndTimes="urn:Exslt.ExsltDatesAndTimes" xmlns:Exslt.ExsltMath="urn:Exslt.ExsltMath" xmlns:Exslt.ExsltRegularExpressions="urn:Exslt.ExsltRegularExpressions" xmlns:Exslt.ExsltStrings="urn:Exslt.ExsltStrings" xmlns:Exslt.ExsltSets="urn:Exslt.ExsltSets"
    exclude-result-prefixes="msxml umbraco.library Exslt.ExsltCommon Exslt.ExsltDatesAndTimes Exslt.ExsltMath Exslt.ExsltRegularExpressions Exslt.ExsltStrings Exslt.ExsltSets ">


<xsl:output method="xml" omit-xml-declaration="yes"/>

<xsl:param name="currentPage"/>

<xsl:variable name="minLevel" select="0"/>

<xsl:template match="/">

<!-- start writing XSLT -->

 <xsl:if test="$currentPage/@level &gt; $minLevel">
    <div id="BreadCrumb">
      <ul>
        <xsl:for-each select="$currentPage/ancestor::node [@level &gt; $minLevel and string(data [@alias='umbracoNaviHide']) != '1']">
         
            <a href="{umbraco.library:NiceUrl(@id)}">
              <xsl:value-of select="@nodeName"/>
            </a>
          &gt;
        </xsl:for-each>
        <!-- print currentpage -->
          <xsl:value-of select="$currentPage/@nodeName"/>
      </ul>
    </div>
    </xsl:if>

</xsl:template>

</xsl:stylesheet>

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