[xsl] How to sort based on the number of child elements?

["Roger L Costello costello@xxxxxxxxx" <xsl-list-service@xxxxxxxxxxxxxxxxxxxxxx>]

Hi Folks,

I have an XML Schema that contains a xs:choice. I want to sort the branches of
the choice, in longest-to-shortest order.

Here is a xs:choice with two branches:

<xs:choice>
    <xs:element name="MilitaryDayTime">
        <xs:complexType>
            <xs:sequence>
                <xs:element name="Day" type="xs:string"/>
                <xs:element name="HourTime" type="xs:string"/>
                <xs:element name="MinuteTime" type="xs:string"/>
                <xs:element name="TimeZone" type="xs:string"/>
            </xs:sequence>
        </xs:complexType>
    </xs:element>
    <xs:element name="DateTimeGroup">
        <xs:complexType>
            <xs:sequence>
                <xs:element name="Day" type="xs:string"/>
                <xs:element name="HourTime" type="xs:string"/>
                <xs:element name="MinuteTime" type="xs:string"/>
                <xs:element name="TimeZone" type="xs:string"/>
                <xs:element name="MonthName" type="xs:string"/>
                <xs:element name="Year" type="xs:string"/>
            </xs:sequence>
        </xs:complexType>
    </xs:element>
</xs:choice>

The first branch is an element with 4 child elements. The second branch is an
element with 6 child elements. So sorting the branches longest-to-shortest
will result in reversing the order of the branches.

I cannot use xsl:sort for this, right?

Is there an easy solution to this task?

/Roger