https://blog.csdn.net/summonlight/article/details/56673544
https://blog.csdn.net/martayang/article/details/54862753
/**
Ag其實是很難求的，需要從整體把握
sigma{(g + 1) * Ag} = sigma{Ag} + sigma{g * Ag}，而
1.sigma{Ag}其實是所有情況
2.g * Ag = Ag + Ag + ... + Ag ,將g*Ag分攤到g個格子上，
那麼sigma{g * Ag}其實是每個格子上分攤到的Ag之和
*/
//#include <bits/stdc++.h>
#include<cstdio>
#include<algorithm>
#include<iostream>
#include<set>
#define T__ int T;scanf("%d",&T);while(T--)
using namespace std;
typedef long long LL;
const int maxn=100+5;
const int MOD=1e9+7;
LL fast_pow(LL a,LL b){
LL rst=1;
for(;b;b>>=1,a=a*a%MOD){
if(b&1) rst=rst*a%MOD;
}
return rst;
}
int main()
{
#ifdef _WY
freopen("0905.in","r",stdin);
#endif // _WY
ios::sync_with_stdio(false);
int kase=0;
LL n,m,k;
int T;cin>>T;
while(T--){
cin>>n>>m>>k;
LL ans=fast_pow(k,m*n);
LL x=(n*m%MOD*fast_pow(k,(n-1)*(m-1))%MOD)%MOD;
LL y=0;
for(int i=1;i<=k;++i){
y=(y+fast_pow(i-1,n+m-2)%MOD)%MOD;
}
ans=(ans+x*y%MOD)%MOD;
cout<<"Case #"<<++kase<<": "<<ans<<endl;
}
return 0;
}