#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstring>
#include<algorithm>
#include<queue>
#include<vector>
#include<cmath>
#include<map>
#include<stack>
#include<set>
#include<string>
using namespace std;
typedef 
 
long long  LL;
typedef unsigned long long ULL;
#define MAXN 49
#define MOD 10000007
#define INF 1000000009
const double eps = 1e-9;
//矩陣快速冪
int n, k;
struct Mat
{
    int a[MAXN][MAXN];
    Mat()
    {
        memset(a, 0, sizeof(a));
    }
    Mat operator *(const Mat& rhs)
    {
        Mat ret;
        for (int
 
 i = 0; i < n; i++)
        {
            for (int j = 0; j < n; j++)
            {
                if (a[i][j])
                {
                    for (int t = 0; t < n; t++)
                    {
                        ret.a[i][t] = (ret.a[i][t] + a[i][j] * rhs.a[j][t])%10;
                    }
                }
            }
        }
        
 
return ret;
    }
    Mat operator +(const Mat& rhs)
    {
        Mat ret;
        for (int i = 0; i < n; i++)
        {
            for (int j = 0; j < n; j++)
            {
                ret.a[i][j] += (a[i][j] + rhs.a[i][j])%10;
            }
        }
        return ret;
    }
};
Mat e;
Mat fpow(const Mat& m, int b)
{
    Mat ans, tmp = m;
    for (int i = 0; i < n; i++)
        ans.a[i][i] = 1;
    while (b != 0)
    {
        if (b & 1)
            ans = tmp*ans;
        tmp = tmp * tmp;
        b >>= 1;
    }
    return ans;
}
Mat sum(const Mat& m, int k)
{
    if (k == 1) return m;
    else if (k % 2 == 0)
    {
        return (e + fpow(m, k / 2)) * sum(m, k / 2);
    }
    else if (k % 2 == 1)
    {
        Mat tmp = fpow(m, k / 2 + 1);
        return    (e + tmp)*sum(m, k / 2) + tmp;
    }
}

int main()   
{
    while (scanf("%d%d", &n,&k), n)
    {
        for (int i = 0; i < n; i++)
            e.a[i][i] = 1;
        Mat M;
        for (int i = 0; i < n; i++)
        {
            for (int j = 0; j < n; j++)
            {
                scanf("%d", &M.a[i][j]);
                M.a[i][j] %= 10;
            }
        }
        M = sum(M, k);
        for (int i = 0; i < n; i++)
        {
            printf("%d", M.a[i][0]);
            for (int j = 1; j < n; j++)
            {
                printf("% d", M.a[i][j]);
            }
            printf("\n");
        }
        printf("\n");
    }
}

Power of Matrix 等比數列求和 矩陣版！