Rikka and Yuta are playing a simple matrix game. At the beginning of the game, Rikka shows an n×m integer matrix A. And then Yuta needs to choose an integer in [1,n], Rikka needs to choose an integer in [1,m]. Let i be Yuta's number and j be Rikka's number, the final score of the game is Ai,j.

In the remaining part of this statement, we use (i,j) to denote the strategy of Yuta and Rikka.

For example, when n=m=3 and matrix A is

If the strategy is (1,2), the score will be 2; if the strategy is (2,2), the score will be 4.

A pure strategy Nash equilibrium of this game is a strategy (x,y) which satisfies neither Rikka nor Yuta can make the score higher by changing his(her) strategy unilaterally. Formally, (x,y) is a Nash equilibrium if and only if:

In the previous example, there are two pure strategy Nash equilibriums: (3,1) and (2,2).

To make the game more interesting, Rikka wants to construct a matrix A for this game which satisfies the following conditions:
1. Each integer in [1,nm] occurs exactly once in A.
2. The game has at most one pure strategy Nash equilibriums.

Now, Rikka wants you to count the number of matrixes with size n×m which satisfy the conditions.

The first line of each testcase contains three numbers n,m and K(1≤n,m≤80,1≤K≤109).

The input guarantees that there are at most 3 testcases with max(n,m)>50.

在一个矩阵中，如果仅有一个数同时是所在行所在列最大值，那么这个数满足纳什均衡。

构造一个n*m的矩阵，里面填入[1,n*m]互不相同的数字，求有多少种构造方案。

由题可得，满足纳什均衡的数一定是最大值n*m。因此我们可以从大往小依次将数填入矩阵，小数依附于大数的行或列，由此产生的三种行为：

1.所在列有大数，行+1 数+1

2.所在行有大数，列+1 数+1

3.所在行列都有大数，位于交界处，数+1

dp三种状态[占有行数][占有列数][占有个数]进行求解。因为本题数据很强，所以需要以下优化：

1.尽可能得减少取模次数

2.注意状态与遍历顺序保持一致

//记忆化搜索

#include<bits/stdc++.h>

using namespace std;

typedef long long ll;

ll mod,n,m;

ll dp[][][];

ll dfs(ll x,ll y,ll z){

    if(dp[x][y][z]>-) return dp[x][y][z];

    ll tmp=;

    if(x<n) tmp+=y*(n-x)*dfs(x+,y,z+)%mod;

    if(y<m) tmp+=x*(m-y)*dfs(x,y+,z+)%mod;

    if(x*y>z) tmp+=(x*y-z)*dfs(x,y,z+)%mod;

    return dp[x][y][z]=tmp;

}

int main()

{

    ll t;

    scanf("%lld",&t);

    while(t--){

        memset(dp,-,sizeof(dp));  //答案有0的情况，因此初始化为-1

        scanf("%lld%lld%lld",&n,&m,&mod);

        dp[n][m][n*m]=;

        ll ans=n*m*dfs(,,)%mod;

        printf("%lld\n",ans);

    }

}

//dp

#include<bits/stdc++.h>

#define MAX 82

#define INF 0x3f3f3f3f

using namespace std;

typedef long long ll;

ll dp[MAX][MAX][];    //注意顺序

int main()

{

    int t,n,m,MOD,i,j,k;

    scanf("%d",&t);

    while(t--){

        scanf("%d%d%d",&n,&m,&MOD);

        memset(dp,,sizeof(dp));

        dp[][][]=n*m;

        for(i=;i<=n;i++){

            for(j=;j<=m;j++){

                for(k=;k<n*m;k++){    //注意顺序

                    dp[i+][j][k+]+=(n-i)*j*dp[i][j][k];

                    if(dp[i+][j][k+]>=MOD) dp[i+][j][k+]%=MOD;

                    dp[i][j+][k+]+=(m-j)*i*dp[i][j][k];

                    if(dp[i][j+][k+]>=MOD) dp[i][j+][k+]%=MOD;

                    if(i*j>k){

                        dp[i][j][k+]+=(i*j-k)*dp[i][j][k];

                        if(dp[i][j][k+]>=MOD) dp[i][j][k+]%=MOD;

                    }

                }

            }

        }

        printf("%lld\n",dp[n][m][n*m]%MOD);

    }

    return ;

}