XSLT position of chosen elements

I'd like to display position of chosen nodes.

xslt:

<xsl:stylesheet version="1.0" xmlns:xsl="http://ift.tt/tCZ8VR">
    <xsl:call-template name="Convert"></xsl:call-template>

    <xsl:template name="Convert">
        <Pos>
            <Type><xsl:value-of select="type"/></Type>
            <LineNumber><xsl:value-of select="position()"/></LineNumber>
            ...
        </Pos>
    </xsl:template>
</xsl:stylesheet>

My xml is:

<OrderPositions>
    <Pos>
         <Type>simple</Type>
         <LineNumber>1</LineNumber>
         ...
    </Pos>
    <Pos>
         <Type>complex</Type>
         <LineNumber>2</LineNumber>
         ...
    </Pos>
    <Pos>
         <Type>simple</Type>
         <LineNumber>3</LineNumber>
         ...
    </Pos>
    <Pos>
         <Type>complex</Type>
         <LineNumber>4</LineNumber> 
         ...
    </Pos>
</OrderPositions>

Trying to exclude "complex" types in xslt using <xsl:choose> + <xsl:when test="type='simple'"> gives LineNumber 1,3,5,7 etc. What is the proper way to choose only "simple" (remove "complex") types and display it positions like 1,2,3,4 etc.?