xquery得到祖先节点的问题

如何得到当前节点的祖先节点？
XML code

<root>
    <students id=1>
        <student>
            <name>Jim</name>
            <age>22</age>
        </student>
        <student>
            <name>Jack</name>
            <age>33</age>
        </student>
    </students>
    <students id=2>
        <student>
            <name>Marry</name>
            <age>21</age>
        </student>
    </students>
</root>

我想得到名字为Marry的student（包含祖先节点）
如：XML code

<root>
    <students id=2>
        <student>
            <name>Marry</name>
            <age>21</age>
        </student>
    </students>
</root>

我写的xquery语句为
for $var in doc("student.xml")/root/students/student
where $var/name="Marry"
return ???
虽然能够得到满足条件的student节点，但是如何得到其祖先节点，从而得到上述结果？

XML code

for $var in doc("student.xml")/root/students/student
where $var/name="Marry"
return <root>{$var/..}</root>

结果（saxonhe运行）
XML code

<?xml version="1.0" encoding="UTF-8"?>
<root>
   <students id="2">
        <student>
            <name>Marry</name>
            <age>21</age>
        </student>
    </students>
</root>