Problem: For each of the K^{NM} ways to write an integer between 1 and K (inclusive) in every square in a square grid with N rows and M columns, find the value defined below, then compute the sum of all those K^{NM} values, modulo D.

* For each of the NM squares, find the minimum among the N+M-1 integers written in the square's row or the square's column. The value defined for the grid is the product of all these NM values.

Constraints

* 1 \leq N,M,K \leq 100
* 10^8 \leq D \leq 10^9
* N,M,K, and D are integers.
* D is prime.

Input

Input is given from Standard Input in the following format:


N M K D


Output

Print the sum of the K^{NM} values, modulo D.

Examples

Input

2 2 2 998244353


Output

35


Input

2 3 4 998244353


Output

127090


Input

31 41 59 998244353


Output

827794103

Solution (in CPP):

Solution:
// #pragma GCC target("avx")  // CPU 処理並列化
// #pragma GCC optimize("O3")  // CPU 処理並列化
// #pragma GCC optimize("unroll-loops")  // 条件処理の呼び出しを減らす
// #define BEGIN_STACK_EXTEND(size) void * stack_extend_memory_ = malloc(size);void * stack_extend_origin_memory_;char * stack_extend_dummy_memory_ = (char*)alloca((1+(int)(((long long)stack_extend_memory_)&127))*16);*stack_extend_dummy_memory_ = 0;asm volatile("mov %%rsp, %%rbx\nmov %%rax, %%rsp":"=b"(stack_extend_origin_memory_):"a"((char*)stack_extend_memory_+(size)-1024));
// #define END_STACK_EXTEND asm volatile("mov %%rax, %%rsp"::"a"(stack_extend_origin_memory_));free(stack_extend_memory_);

#include<stdio.h>
#include<math.h>
#include<algorithm>
#include<queue>
#include<deque>
#include<stack>
#include<string>
#include<string.h>
#include<vector>
#include<set>
#include<map>
#include<bitset>
#include<stdlib.h>
#include<cassert>
#include<time.h>
#include<bitset>
#include<numeric>
#include<unordered_set>
#include<unordered_map>
#include<complex>
using namespace std;
// const long long mod=998244353;
// const long long inf=mod*mod;
// const long long d2=(mod+1)/2;
long long mod;
const long double EPS=1e-9;
const long double INF=1e+10;
const long double PI=acos(-1.0);
const int C_SIZE = 3100000;
const int UF_SIZE = 3100000;
namespace{
	long long fact[C_SIZE];
	long long finv[C_SIZE];
	long long inv[C_SIZE];
	inline long long Comb(int a,int b){
	 	if(a<b||b<0)return 0;
	 	return fact[a]*finv[b]%mod*finv[a-b]%mod;
	}
	void init_C(int n){
		fact[0]=finv[0]=inv[1]=1;
		for(int i=2;i<n;i++){
			inv[i]=(mod-(mod/i)*inv[mod%i]%mod)%mod;
		}
		for(int i=1;i<n;i++){
			fact[i]=fact[i-1]*i%mod;
			finv[i]=finv[i-1]*inv[i]%mod;
		}
	}
	long long pw(long long a,long long b){
		if(a<0LL)return 0;
		if(b<0LL)return 0;
		long long ret=1;
		while(b){
			if(b%2)ret=ret*a%mod;
			a=a*a%mod;
			b/=2;
		}
		return ret;
	}
	long long pw_mod(long long a,long long b,long long M){
		if(a<0LL)return 0;
		if(b<0LL)return 0;
		long long ret=1;
		while(b){
			if(b%2)ret=ret*a%M;
			a=a*a%M;
			b/=2;
		}
		return ret;
	}
	int pw_mod_int(int a,int b,int M){
		if(a<0)return 0;
		if(b<0)return 0;
		int ret=1;
		while(b){
			if(b%2)ret=(long long)ret*a%M;
			a=(long long)a*a%M;
			b/=2;
		}
		return ret;
	}
	int ABS(int a){return max(a,-a);}
	long long ABS(long long a){return max(a,-a);}
	double ABS(double a){return max(a,-a);}
	int sig(double r) { return (r < -EPS) ? -1 : (r > +EPS) ? +1 : 0; }
	int UF[UF_SIZE];
	void init_UF(int n){
		for(int i=0;i<n;i++)UF[i]=-1;
	}
	int FIND(int a){
		if(UF[a]<0)return a;
		return UF[a]=FIND(UF[a]);
	}
	void UNION(int a,int b){
		a=FIND(a);b=FIND(b);if(a==b)return;
		// if(UF[a]>UF[b])swap(a,b);
		UF[a]+=UF[b];UF[b]=a;
	}
}
// ここから編集しろ
long long dp[110][2][110][110];
long long P[110][11000];
long long C[110][110];
int main(){
	int a,b,c;
	scanf("%d%d%d%lld",&a,&b,&c,&mod);
	init_C(210000);
	for(int i=0;i<110;i++){
		P[i][0]=1;
		for(int j=1;j<11000;j++){
			P[i][j]=P[i][j-1]*i%mod;
		}
	}
	C[0][0]=1;
	for(int i=0;i<105;i++){
		for(int j=0;j<=i;j++){
			C[i+1][j]=(C[i+1][j]+C[i][j])%mod;
			C[i+1][j+1]=(C[i+1][j+1]+C[i][j])%mod;
		}
	}
	dp[0][0][0][0]=1;
	for(int i=0;i<c;i++){
		for(int j=0;j<2;j++){
			for(int k=0;k<=a;k++){
				for(int l=0;l<=b;l++){
					if(!dp[i][j][k][l])continue;
					// printf("%d %d %d %d: %lld\n",i,j,k,l,dp[i][j][k][l]);
					if(j==0){
						long long cur=1;
						for(int m=k;m<=a;m++){
							long long ks=cur;

							ks=ks*C[a-k][a-m]%mod;						
							dp[i][j+1][m][l]=(dp[i][j+1][m][l]+ks*dp[i][j][k][l])%mod;

							cur=cur*(P[i+1][b-l]+mod-P[i][b-l])%mod;
							cur=cur*(P[c-i][l])%mod;
						}
					}else{
						long long cur=1;
						for(int m=l;m<=b;m++){
							long long ks=cur;
							ks=ks*C[b-l][b-m]%mod;
							dp[i+1][0][k][m]=(dp[i+1][0][k][m]+ks*dp[i][j][k][l])%mod;

							cur=cur*(P[i+1][a-k])%mod;
							cur=cur*(P[c-i][k]+mod-P[c-1-i][k])%mod;
						}
					}
				}
			}
		}
	}
	printf("%lld\n",dp[c][0][a][b]);
}