Get the title of file and not file name in content query web part?

Question

What do I use instead of

       <xsl:variable name="DisplayTitle">
            <xsl:call-template name="OuterTemplate.GetTitle">
                <xsl:with-param name="Title" select="''"/>
                <xsl:with-param name="UrlColumnName" select="'LinkUrl'"/>
                <xsl:with-param name="UseFileName" select="1"/>
            </xsl:call-template>
       </xsl:variable>

<xsl:value-of select="$DisplayTitle"/>

in the XSL template to get the title of a file rather than its file name?

Answer 1

Edited out of question:

I have obviously not had enough coffee... I replaced the variable above with:

<xsl:variable name="DisplayTitle">
    <xsl:call-template name="OuterTemplate.GetTitle">
        <xsl:with-param name="Title" select="@Title"/>
        <xsl:with-param name="UrlColumnName" select="'LinkUrl'"/>
    </xsl:call-template>
</xsl:variable>