Shanto is learning how to power up numbers and he found an efficient way to find k^th power of a matrix. He was quite happy with his discovery. Suddenly his sister Natasha came to him and asked him to find the summation of the powers. To be specific his sister gave the following problem.

Let A be an n x n matrix. We define A^k = A * A * ... * A (k times). Here, * denotes the usual matrix multiplication. You are to write a program that computes the matrix A + A² + A³ + ... + A^k.

Shanto smiled and thought that it would be an easy one. But after a while he found that it‘s tough for him. Can you help him?

Input

Input starts with an integer T (≤ 20), denoting the number of test cases.

Each case starts with two integers n (1 ≤ n ≤ 30) and k (1 ≤ k ≤ 10⁹). Each of the next n lines will contain n non-negative integers (not greater than 10).

Output

For each case, print the case number and the result matrix. For each cell, just print the last digit. See the samples for more details.

  1 #include<stdio.h>
  2 #include<algorithm>
  3 #include<iostream>
  4 #include<stdlib.h>
  5 #include<string.h>
  6 #include<math.h>
  7 #include<queue>
  8 using namespace std;
  9 typedef long long LL;
 10 typedef struct node
 11 {
 12     int  m[70][70];
 13     node()
 14     {
 15         memset(m,0,sizeof(m));
 16     }
 17 } maxtr;
 18 int  ans[70][70];
 19 void E(node *nn,int n);
 20 maxtr ju(int n);
 21 maxtr quick(node ju,int n,int m);
 22 int main(void)
 23 {
 24     int i,j,k;
 25     int n,m;
 26     int s;
 27     cin>>k;
 28     for(s=1; s<=k; s++)
 29     {
 30         scanf("%d %d",&n,&m);
 31         for(i=0; i<n; i++)
 32         {
 33             for(j=0; j<n; j++)
 34             {
 35                 scanf("%d",&ans[i][j]);
 36                 ans[i][j]%=10;
 37             }
 38         }node aa;aa=ju(n);
 39         aa=quick(aa,n,m);printf("Case %d:\n",s);
 40         for(i=0;i<n;i++)
 41         {
 42             for(j=n;j<2*n;j++)
 43             {
 44                 printf("%d",aa.m[i][j]);
 45             }printf("\n");
 46         }
 47     }return 0;
 48 }
 49 void E(node *nn,int n)
 50 {
 51     int i,j,k;
 52     for(i=0; i<n; i++)
 53     {
 54         for(j=0; j<n; j++)
 55         {
 56             if(i==j)
 57                 nn->m[i][j]=1;
 58             else nn->m[i][j]=0;
 59         }
 60     }
 61 }
 62 maxtr ju(int n)
 63 {
 64     int i,j,k;
 65     maxtr nn;
 66     for(i=0; i<n; i++)
 67     {
 68         for(j=0; j<n; j++)
 69         {
 70             nn.m[i][j]=ans[i][j];
 71         }
 72     }
 73     for(i=0; i<n; i++)
 74     {
 75         for(j=n; j<2*n; j++)
 76         {
 77             nn.m[i][j]=ans[i][j-n];
 78         }
 79     }
 80     node cc;
 81     E(&cc,n);
 82     for(i=n; i<2*n; i++)
 83     {
 84         for(j=n; j<2*n; j++)
 85         {
 86             nn.m[i][j]=cc.m[i-n][j-n];
 87         }
 88     }return nn;
 89 }
 90 maxtr quick(node ju,int n,int m)
 91 {   node ee;
 92
 93     E(&ee,2*n);
 94     int i,j,k;
 95     int s;
 96     while(m)
 97     {
 98         if(m&1)
 99         {
100             node cc;
101             for(i=0; i<2*n; i++)
102             {
103                 for(j=0; j<2*n; j++)
104                 {
105                     for(s=0; s<2*n; s++)
106                     {
107                         cc.m[i][j]=(ju.m[i][s]*ee.m[s][j]+cc.m[i][j])%10;
108                     }
109                 }
110             }
111             ee=cc;
112         }
113         node cc;
114         for(i=0; i<2*n; i++)
115         {
116             for(j=0; j<2*n; j++)
117             {
118                 for(s=0; s<2*n; s++)
119                 {
120                     cc.m[i][j]=(ju.m[i][s]*ju.m[s][j]+cc.m[i][j])%10;
121                 }
122             }
123         }
124         ju=cc;
125         m/=2;
126     }
127     return ee;
128 }