Literary and artistic calculation Ji of provincial election and professional training

“ Fighting for three weeks , Build a computer ”. Small W Answer the call , It took three weeks to build a TV station .

Literary computing has more artistic cells than ordinary computers . An ordinary computer can calculate the number of spanning trees of a labeled complete graph , And literary computing can calculate the number of spanning trees of a labeled complete bipartite graph .

More specifically , The number of points on a given side is n, The points on the other side are m, share n*m Labeled complete bipartite graph with edges K_{n,m}, Calculation can quickly calculate the number of spanning trees .

Small W I don't know if Ji's calculation is right , Can you help him ？

This is actually a favorite ** Counting problems
The solution to this problem is Prufer Sequence
We know that we are looking for a tree Prufer The last two points are left in the sequence
They must be in two partitioned sets
At this time, it means n-1 And m-1 Of them were selected
Then according to the Prufer The nature of
How many times a point appears is related to his degree
Then you can choose any number of times this point appears
by ：$N^{M-1}\ast M^{N-1}$

#include<iostream>
#include<cstring>
#include<cstdio>
#include<algorithm>
#include<cmath>
using namespace std;
typedef long long LL;
LL mod,n,m;
LL mul(LL A,LL B){
	LL ret=0;
	while(B){
		if(B&1)ret=(ret+A)%mod;
		A=(A+A)%mod;
		B=B>>1;
	}
	return ret;
}
LL Quick_Pow(LL x,LL k){
	LL ret=1;
	while(k){
		if(k&1)ret=mul(ret,x)%mod;
		x=mul(x,x)%mod;
		k=k>>1;
	}
	return ret;
}
int main(){
	cin>>n>>m>>mod;
	cout<<mul(Quick_Pow(n,m-1),Quick_Pow(m,n-1));
}