#include <bits/stdc++.h>
using namespace std;
//矩阵快速幂
typedef long long LL;
const int MOD = 1e9 + 7;
const int N = 110;
LL n, k; //本题数据范围很大，用int直接wa哭了

//矩阵声明
struct JZ {
    LL m[N][N];
} A, res, base;

//矩阵乘法
inline JZ mul(JZ A, JZ B) {
    JZ C;
    memset(C.m, 0, sizeof(C.m));
    for (LL i = 0; i < n; i++)
        for (LL j = 0; j < n; j++)
            for (LL k = 0; k < n; k++) {
                C.m[i][j] += (A.m[i][k] % MOD) * (B.m[k][j] % MOD);
                C.m[i][j] %= MOD;
            }
    return C;
}

void qmi() {
    //将结果矩阵初始化为单位矩阵
    memset(res.m, 0, sizeof res.m);
    for (int i = 0; i < n; i++) res.m[i][i] = 1;
    //其实就是把整数快速幂修改为矩阵快速幂
    while (k) {                       //二进制快速幂
        if (k & 1) res = mul(res, A); // 联想一下整数快速幂
        A = mul(A, A);                // base 翻倍
        k >>= 1;
    }
}
int main() {
    cin >> n >> k;
    //输入原始矩阵
    for (int i = 0; i < n; i++)
        for (int j = 0; j < n; j++)
            cin >> A.m[i][j];

    //计算矩阵快速幂
    qmi();

    //输出
    for (int i = 0; i < n; i++) {
        for (int j = 0; j < n; j++)
            cout << res.m[i][j] << " ";
        cout << endl;
    }
    return 0;
}