I have one question about the xslt and the navigation menu.

If i have a structure like this:

- Home

          - About us

          - Our projects

- Login

And i have this code:

<?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"
    exclude-result-prefixes="msxml umbraco.library">


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

<xsl:param name="currentPage"/>

<xsl:template match="/">

<!-- start writing XSLT -->

<ul id="nav">
<xsl:for-each select="$currentPage/ancestor::root/node [string(./data [@alias='umbracoNaviHide']) != '1']">
    <li>
        <xsl:if test="$currentPage/@id=current()/@id">
            <xsl:attribute name="class">Selected</xsl:attribute>
        </xsl:if>
        <a href="{umbraco.library:NiceUrl(@id)}">
            <xsl:attribute name="title"><xsl:value-of select="@nodeName" /></xsl:attribute>
            <xsl:value-of select="@nodeName" />
        </a>
    </li>
</xsl:for-each>

<xsl:for-each select="$currentPage/ancestor-or-self::node [@level=1]/node [string(./data [@alias='umbracoNaviHide']) != '2']">
    <li>
        <xsl:if test="$currentPage/ancestor-or-self::node/@id = current()/@id">
            <xsl:attribute name="class">Selected</xsl:attribute>
        </xsl:if>
        <a href="{umbraco.library:NiceUrl(@id)}">
            <xsl:attribute name="title"><xsl:value-of select="@nodeName" /></xsl:attribute>
            <xsl:value-of select="@nodeName" />
        </a>
    </li>
</xsl:for-each>
</ul>


</xsl:template>

</xsl:stylesheet>

Of course the navigation looks like this:

Home Login About us Our projects

But how can i have just navigation for "Home", so without the "Login".

Tnx