One day, Alice and Bob felt bored again, Bob knows Alice is a girl who loves math and is just learning something about matrix, so he decided to make a crazy problem for her. 

Bob has a six-faced dice which has numbers 0, 1, 2, 3, 4 and 5 on each face. At first, he will choose a number N (4 <= N <= 1000), and for N times, he keeps throwing his dice for K times (2 <=K <= 6) and writes down its number on the top face to make an N*K matrix A, in which each element is not less than 0 and not greater than 5. Then he does similar thing again with a bit difference: he keeps throwing his dice for N times and each time repeat it for K times to write down a K*N matrix B, in which each element is not less than 0 and not greater than 5. With the two matrix A and B formed, Alice’s task is to perform the following 4-step calculation. 

Step 1: Calculate a new N*N matrix C = A*B. 
Step 2: Calculate M = C^(N*N). 
Step 3: For each element x in M, calculate x % 6. All the remainders form a new matrix M’. 
Step 4: Calculate the sum of all the elements in M’. 

Bob just made this problem for kidding but he sees Alice taking it serious, so he also wonders what the answer is. And then Bob turn to you for help because he is not good at math. InputThe input contains several test cases. Each test case starts with two integer N and K, indicating the numbers N and K described above. Then N lines follow, and each line has K integers between 0 and 5, representing matrix A. Then K lines follow, and each line has N integers between 0 and 5, representing matrix B. 

The end of input is indicated by N = K = 0.OutputFor each case, output the sum of all the elements in M’ in a line.Sample Input

4 2
5 5
4 4
5 4
0 0
4 2 5 5
1 3 1 5
6 3
1 2 3
0 3 0
2 3 4
4 3 2
2 5 5
0 5 0
3 4 5 1 1 0
5 3 2 3 3 2
3 1 5 4 5 2
0 0

Sample Output

14
56

AC C++:

#include <iostream>  
#include <cstdio>  
#include <cstring>  
//#define DEBUG
using namespace std;  
int n,k;  

struct matrix  
{  
    int map[7][7];  
    matrix()
    {
    	memset(map,0,sizeof(map));
	} 
};
  
matrix mult(matrix a,matrix b)  
{  
    matrix c;  
    for(int i=1;i<=k;i++)  
        for(int j=1;j<=k;j++)  
        {  
            for(int m=1;m<=k;m++)    
                c.map[i][j]=(c.map[i][j]+a.map[i][m]*b.map[m][j]);
            c.map[i][j]%=6;  
        }  
    
    return c;  
}  

matrix pow_mult(matrix a,int x)  
{  
    matrix c;  
    for(int i=1;i<=k;i++)  
        for(int j=1;j<=k;j++)  
            c.map[i][j]=(i==j);  
    
    while(x)
    {  
        if(x&1)c=mult(c,a);
		x>>=1;  
        a=mult(a,a);  
    }  
    return c;  
}  

int a[1005][7],b[7][1005],final[7][1005],mid[7][7],last[1005][1005],sum;  
/*
發現使用快速冪矩陣之後依然超時，非常絕望，後來就把 （A*B）^(n*n-1)拆開變成 A*(A*B)^(n*n-1)*B，
然後拆出來的一個A和一個B用 陣列 而非 結構體 ，直接在 main() 裡面計算 而非 傳遞到子函式，以此節省時間來AC程式碼。 
但是一定要注意矩陣乘法是不可以顛倒左右兩個矩陣的次序的，
因此最後乘以A 然後再 乘以B的時候一定要保證順序和迴圈界限不要寫錯。

*/
int main()  
{  
#ifdef DEBUG	
	freopen("try.txt","r",stdin);
#endif 
    while(scanf("%d%d",&n,&k) && (n!=0)&&(k!=0))   
    {  
      for(int i=1;i<=n;i++)  
        for(int j=1;j<=k;j++)  
        	scanf("%d",&a[i][j]);     
      
      for(int i=1;i<=k;i++)  
        for(int j=1;j<=n;j++)  
	        scanf("%d",&b[i][j]);     

      memset(mid,0,sizeof(mid));  
      ;
      for(int i=1;i<=k;i++)    
        for(int j=1;j<=k;j++)  
        {  
           for(int m=1;m<=n;m++)    
             	mid[i][j]=mid[i][j]+b[i][m]*a[m][j];              
           mid[i][j]%=6;      
        }  
  
      matrix ans;  
      for(int i=1;i<=k;i++)  
        for(int j=1;j<=k;j++)  
            ans.map[i][j]=mid[i][j];  
      
#ifdef DEBUG      
      	for(int i=1;i<=k;i++)  
      	{  
	        for(int j=1;j<=n;j++)  
	        {
   		     	cout<<ans.map[i][j]<<' ';
			}
			cout<<endl;
		}
#endif
      
      ans=pow_mult(ans,n*n-1);  
      memset(final,0,sizeof(final));  
      for(int i=1;i<=k;i++)  
      {  
        for(int j=1;j<=n;j++)  
        {  
            for(int m=1;m<=k;m++)  
            {  
                final[i][j]=final[i][j]+ans.map[i][m]*b[m][j];  
            }  
            final[i][j]%=6;  
        }  
      }  
      memset(last,0,sizeof(last));  
      for(int i=1;i<=n;i++)  
      {  
        for(int j=1;j<=n;j++)  
        {  
            for(int m=1;m<=k;m++)  
            {  
                last[i][j]=(last[i][j]+a[i][m]*final[m][j]);  
            }  
            last[i][j]%=6;  
        }  
      }  
      sum=0;  
      for(int i=1;i<=n;i++)  
      {  
        for(int j=1;j<=n;j++)  
        {  
            sum+=last[i][j];  
        }  
      }  
      cout<<sum<<endl;  
    }  
    return 0;  
}