for-each: get every 3rd item (split into 3 columns)

I have the following nodes in xsl:

<foo>
    <bar>1</bar>
    <bar>2</bar>    
    <bar>3</bar>
    <bar>4</bar>        
    <bar>5</bar>
    <bar>6</bar>    
    <bar>7</bar>
    <bar>8</bar>            
    <bar>9</bar>
</foo>

And would like to turn it into the following html:

<ul class="one">
    <li>1</li>
    <li>4</li>
    <li>7</li>
</ul>
<ul class="two">
    <li>2</li>
    <li>5</li>
    <li>8</li>
</ul>
<ul class="three">
    <li>3</li>
    <li>6</li>
    <li>9</li>
</ul>

Having a hard time figuring out how to loop and get开发者_运维知识库 each third item, would like to do something like this:

<ul class="one">
<xsl:for-each select="exlt:node-set($blah)/foo/bar[X1]">
    <li><xsl:value-of select="node()"/></li>
</xsl:for-each>
</ul>

<ul class="two">
<xsl:for-each select="exlt:node-set($blah)/foo/bar[X2]">
    <li><xsl:value-of select="node()"/></li>
</xsl:for-each>
</ul>

<ul class="three">
<xsl:for-each select="exlt:node-set($blah)/foo/bar[X3]">
    <li><xsl:value-of select="node()"/></li>
</xsl:for-each>
</ul>   

Where:
X1 = Every third item starting from position 1
X2 = Every third item starting from position 2
X3 = Every third item starting from position 3

Might need to use last(), but can't quite get that working correctly.

In XPath, the condition will be:

not(position() mod 3)

or

position() mod 3 = 0

I don't see why you can't use op:mod.

EDIT: About new question, just substract the offset. So:

X1:

position() mod 3 = 1

X2:

position() mod 3 = 2

X3:

position() mod 3 = 0

EDIT 2: Now I understand your question.

Try [position() mod 3 = 1]