[xsl] replace a string that is in between two specific chars in XSLT 1.0

["Prady Prady prady.chin@xxxxxxxxx" <xsl-list-service@xxxxxxxxxxxxxxxxxxxxxx>]

Team,


Can somebody help me to replace string that is in between* # and ;* with
blank. I need to include this is oracle bpel which only supports xslt 1.0

From:
<IpsCollection xmlns="http://xmlns.oracle.com/aa";>
<Ips>
 * <Ip>q1#11.11.11.111;q2#22.22.22.22</Ip>*
</Ips>
 </IpsCollection>
To:
* <v2:Ops xmlns:v2="http://xmlns.ieee.org/V2 <http://xmlns.ieee.org/V2>">*
<v2:nums>q1;q2</v2:nums>
 </v2:Ops>


Thank you very much for your help!