File System Audit

[SlowYaRoll]

I've been working on this for a while and can't seem to get it right. I'm working on a script that will eventually be run as a cron job. From Linux I'm mounting a remote Windows share as /media/windows_share The share I'm mounting is for user's personal drive. Some people seem to have a bunch of audio files. I can search for them using

find /media/windows_share -iname "*.mp3" -o -iname "*.wma" -o -iname "*.m4a"

I've even pipped that in to

du -h

The output results in tons and tons of full paths to files. The folder structure is /media/windows_share/department_name/user_name for example:
/media/windows_share/finance/john doe

When I run the script, I'd like to it to NOT show me the full path of all the music files but instead show /department/user_name and how much disk space the audio files they're storing are using. *please note that the user names are full first name and full last name separate by a space. So far I've managed to get the a script that will give me just the department name and user name, but I can't get it to give me the disk space usage. I also get a dc: stack empty message

Here's what I've got so far

#!/bin/bash
mtp="/media/windows_share"
>found.files
ls $mtp|while read dept;do
        ls $mtp/$dept|while read user;do
                echo $dept/$user;
                find "$mtp/$dept/$user" -maxdepth 20 -iname "*.wma" -o -iname "*.mp3" -o -iname "*.m4a"|while read files;do
                        count=`du -h "$files"|awk '{print $1}'|grep "M"|awk -F "M" '{print $1}'`;
                        echo -n "$count+">>found.files;
                done
        echo -n "p">>found.files;
        dc -f found.files;
        echo;
        >found.files;
        done
done

Any help would be appreciated