Having issue when there are identical keys in nodeset formed - xslt-1.0

Question

My input xml is as below, I am facing issue when there are multiple identical keys present in a nodeset, its not forming proper output structure. I am using keys to form the nodeset by using ref attribute. I am making use of these <SL id="L1S1"> and <pit ref="L1S1"> to form nodesets. Id attribute of SL node(i.e <SL id="L1S1">) is formed using "L1" in <L Id="L1">. Also ref attribute of pit node(i.e <pit ref="L1S1">) is formed using "L1" in <L Id="L1">.
This solution works fine if I do not have multiple pit nodes with same reference number. I am using xslt1.0, and I have issue when multiple identical keys are found in nodeset.
Input xml as below
<root>
<L Id="L1">
<test>ed</test>
<SL id="L1S1">
<check>
<AId>1</AId>
</check>
</SL>
<SL id="L1S2">
<check>
<AId>2</AId>
</check>
</SL>
</L>
<cp>
<current>
<Amt>20154.00</Amt>
</current>
<pi>
<pit ref="L1S1">
<value>123</value>
</pit>
<pit ref="L1S1">
<value>234</value>
</pit>
<pit ref="L1S2">
<value>1232</value>
</pit>
</pi>
</cp>
</root>
Expected output should be:
<root>
<L Id="L1">
<SL id="L1S1">
<check>
<AId>1</AId>
</check>
</SL>
<pit ref="L1S1">
<value>123</value>
</pit>
</L>
<L Id="L1">
<SL id="L1S1">
<check>
<AId>1</AId>
</check>
</SL>
<pit ref="L1S1">
<value>234</value>
</pit>
</L>
<L Id="L1">
<SL id="L1S2">
<check>
<AId>2</AId>
</check>
</SL>
<pit ref="L1S2">
<value>1232</value>
</pit>
</L>
</root>
<xsl:key name="pit-SL" match="pit" use="#ref" />
<xsl:key name="pit-L" match="pit" use="substring(#ref,1,2)" />
<xsl:template match="#*|node()">
<xsl:copy>
<xsl:apply-templates select="#*|node()"/>
</xsl:copy>
</xsl:template>
<xsl:template match="L[SL]">
<xsl:apply-templates select="SL"/>
</xsl:template>
<xsl:template match="SL">
<L>
<xsl:copy-of select="parent::L/#Id"/>
<xsl:copy>
<xsl:copy-of select="#id"/>
<xsl:apply-templates select="node()"/>
<xsl:copy-of select="key('pit-SL',#id)"/>
</xsl:copy>
</L>
</xsl:template>
<xsl:template match="L[not(SL)]">
<xsl:apply-templates select="key('pit-L',#Id)">
<xsl:with-param name="L" select="."/>
</xsl:apply-templates>
</xsl:template>
<xsl:template match="pit">
<xsl:param name="L"/>
<L>
<xsl:copy-of select="$L/#Id"/>
<xsl:copy-of select="."/>
</L>
</xsl:template>
<xsl:template match="cp"/>

Answer 1

AFAICT, you could do:
XSLT 1.0
<xsl:stylesheet version="1.0"
xmlns:xsl="http://www.w3.org/1999/XSL/Transform">
<xsl:output method="xml" version="1.0" encoding="UTF-8" indent="yes"/>
<xsl:strip-space elements="*"/>
<xsl:key name="pit" match="pit" use="#ref" />
<xsl:template match="/root">
<xsl:copy>
<xsl:for-each select="L">
<xsl:variable name="Id" select="#Id" />
<xsl:for-each select="SL">
<xsl:variable name="SL" select="." />
<xsl:for-each select="key('pit', #id)">
<L>
<xsl:copy-of select="$Id"/>
<xsl:copy-of select="$SL"/>
<xsl:copy-of select="."/>
</L>
</xsl:for-each>
</xsl:for-each>
</xsl:for-each>
</xsl:copy>
</xsl:template>
</xsl:stylesheet>