cf111D Petya and Coloring 组合数学，二项式反演

http://codeforces.com/contest/111/problem/D

Little Petya loves counting. He wants to count the number of ways to paint a rectangular checkered board of size n × m (n rows, m columns) in k colors. Besides, the coloring should have the following property: for any vertical line that passes along the grid lines and divides the board in two non-empty parts the number of distinct colors in both these parts should be the same. Help Petya to count these colorings.

Input

The first line contains space-separated integers n, m and k (1 ≤ n, m ≤ 1000, 1 ≤ k ≤ 10⁶) — the board's vertical and horizontal sizes and the number of colors respectively.

Output

Print the answer to the problem. As the answer can be quite a large number, you should print it modulo 10⁹ + 7 (1000000007).

Examples

Input

2 2 1

Output

1

Input

2 2 2

Output

8

Input

3 2 2

Output

40

题意：
给出一个 n*m 的矩阵，用sum种颜色染色
满足：任意一条竖直线（纵线）把矩阵划分成的2个部分，2个部分的不同的颜色数相同
求方案数
n,m <= 10^3,sum <= 10^6

solution:
注意到n,m的范围不大

显然有以下性质：
1.第1列和第m列的颜色数一定相等
2.2～m-1列的颜色只能从1，m列的颜色的交集中选择

m=1的时候，特殊处理，ans=k^n
m>1的时候，我们只需要考虑1，m列的颜色选择还有交集大小

预处理g[i]表示恰好用i种颜色涂满n个格子的方案数
(此时n是一个常量)

如果用递推式，求g[i]需要O(n^2),求g数组需要O(n^3)，tle
考虑二项式反演求g数组：
设h(i)表示用i种颜色染n个格子的方案数，则h(i) = i^n
g(i)表示恰好用i种颜色染n个格子的方案数，
有：h(y) = sigma(i=0,i<=y)(C(y,i)*g(i))
则：g(y) = sigma(i=0,i<=y)((-1)^(y-i) * C(y,i) * h(i))
　　　　　= sigma(i=0,i<=y)((-1)^(y-i) * C(y,i) * i^n)

这样求g[i]需要O(nlogn),求g数组需要O(n^2*logn)
当然也可以优化到O(n^2)求g数组

主要的预处理部分搞定了，然后就是答案了
退下公式，得到：
ans = sigma(j=0,j<=min(n,sum))C(sum,j) * j^(m*n-2*n) *
　　　 (sigma(i=0,i<=min(n-j,(sum-j)/2))(C(sum-j,i)*C(sum-j-i,i)*g(i+j)^2))

   //File Name: cf111D.cpp
   //Author: long
   //Mail: [e-mail]
   //Created Time: 2016年05月16日 星期一 01时13分24秒

 #include <stdio.h>
 #include <algorithm>
 #include <iostream>
 #include <string.h>
 #include <stdlib.h>
 #include <math.h>

 #define LL long long

 using namespace std;

 const int MAXN =  + ;
 const int MAXM =  + ;
 const int MOD = (int)1e9 + ;

 LL jie[MAXM];
 LL g[MAXN];

 LL qp(LL x,LL y){
     LL res = ;
     while(y){
         if(y & ) res = res * x % MOD;
         x = x * x % MOD;
         y >>= ;
     }
     return res;
 }

 LL get_c(LL x,LL y){
     if(x < y) return ;
     if(x == y || y == ) return ;
     return jie[x] * qp(jie[y] * jie[x - y] % MOD,MOD - ) % MOD;
 }

 void init(int sum,int n){
     jie[] = ;
     for(int i=;i<MAXM;i++)
         jie[i] = jie[i-] * i % MOD;
     int ma = min(sum,n);
     LL now;
     for(int i=;i<=ma;i++){
         for(int k=;k<=i;k++){
             now = get_c(i,k) * qp(k,n) % MOD;
             if((i - k) % )
                 g[i] = (g[i] - now + MOD) % MOD;
             else
                 g[i] = (g[i] + now) % MOD;
         }
     }
 }

 LL solve(int n,int m,int sum){
     if(m == ) return qp(sum,n);
     init(sum,n);
     LL ans = ,now,tmp;
     int ma = min(sum,n);
     for(int j=,ma2;j<=ma;j++){
         now = ;
         ma2 = min(n - j,(sum - j) / );
         for(int i=;i<=ma2;i++){
             (now += jie[sum-j] * qp(jie[sum-j-*i]*jie[i]%MOD*jie[i]%MOD,MOD - ) % MOD
                 * g[i+j] % MOD * g[i+j] % MOD) %= MOD;
         }
         tmp = qp(j,(m - ) * n) % MOD;
         (ans += get_c(sum,j) * tmp % MOD * now % MOD) %= MOD;
         //cout << j << " " <<ans << endl;
     }
     return ans;
 }

 int main(){
     int n,m,k;
     scanf("%d %d %d",&n,&m,&k);
     printf("%d\n",(int)solve(n,m,k));
     return ;
 }