Problem:
Score : 100 points

Problem Statement

There is a simple directed graph G with N vertices, numbered 1, 2, \ldots, N.

For each i and j (1 \leq i, j \leq N), you are given an integer a_{i, j} that represents whether there is a directed edge from Vertex i to j.
If a_{i, j} = 1, there is a directed edge from Vertex i to j; if a_{i, j} = 0, there is not.

Find the number of different directed paths of length K in G, modulo 10^9 + 7.
We will also count a path that traverses the same edge multiple times.

Constraints

All values in input are integers.

1 \leq N \leq 50

1 \leq K \leq 10^{18}

a_{i, j} is 0 or 1.

a_{i, i} = 0

Input

Input is given from Standard Input in the following format:

N K
a_{1, 1} \ldots a_{1, N}
:
a_{N, 1} \ldots a_{N, N}

Output

Print the number of different directed paths of length K in G, modulo 10^9 + 7.

Sample Input 1

4 2
0 1 0 0
0 0 1 1
0 0 0 1
1 0 0 0

Sample Output 1

6

G is drawn in the figure below:

There are six directed paths of length 2:

1 → 2 → 3

1 → 2 → 4

2 → 3 → 4

2 → 4 → 1

3 → 4 → 1

4 → 1 → 2

Sample Input 2

3 3
0 1 0
1 0 1
0 0 0

Sample Output 2

3

G is drawn in the figure below:

There are three directed paths of length 3:

1 → 2 → 1 → 2

2 → 1 → 2 → 1

2 → 1 → 2 → 3

Sample Input 3

6 2
0 0 0 0 0 0
0 0 1 0 0 0
0 0 0 0 0 0
0 0 0 0 1 0
0 0 0 0 0 1
0 0 0 0 0 0

Sample Output 3

1

G is drawn in the figure below:

There is one directed path of length 2:

4 → 5 → 6

Sample Input 4

1 1
0

Sample Output 4

0

Sample Input 5

10 1000000000000000000
0 0 1 1 0 0 0 1 1 0
0 0 0 0 0 1 1 1 0 0
0 1 0 0 0 1 0 1 0 1
1 1 1 0 1 1 0 1 1 0
0 1 1 1 0 1 0 1 1 1
0 0 0 1 0 0 1 0 1 0
0 0 0 1 1 0 0 1 0 1
1 0 0 0 1 0 1 0 0 0
0 0 0 0 0 1 0 0 0 0
1 0 1 1 1 0 1 1 1 0

Sample Output 5

957538352

Be sure to print the count modulo 10^9 + 7.

Solution in C++ (GCC 9.2.1) (judge verdict: Accepted)

#include <bits/stdc++.h>
#define F first
#define S second
using namespace std;
typedef long long LL;
typedef pair<int, int> pii;
const LL INF = 0x3f3f3f3f3f3f3f3fLL;
const LL MOD = 1e9 + 7;
typedef vector<LL> VL;
typedef vector<VL> ML;
ML operator * (const ML &a, const ML &b) {
    int n = a.size(), m = b[0].size();
    int k = a[0].size();
    ML ret(n, VL(m, 0));
    for (int i = 0 ; i < n ; i++)
        for (int j = 0 ; j < m ; j++)
            for (int x = 0 ; x < k ; x++)
                (ret[i][j] += a[i][x] * b[x][j]) %= MOD;
    return ret;
}
ML fpow(auto &A, LL n) {
    if (n == 1) return A;
    auto k = fpow(A, n >> 1);
    if (n & 1)
        return k * k * A;
    else
        return k * k;
}
void marmot0814() {
    int n; LL k; cin >> n >> k;
    ML A(n, VL(n));
    for (auto &vv : A)
        for (auto &v : vv)
            cin >> v;
    ML res = fpow(A, k);
    LL ans = 0;
    for (auto &vv : res)
        for (auto &v : vv)
            (ans += v) %= MOD;
    cout << ans << '\n';
}
int main() {
    ios_base::sync_with_stdio(false); cin.tie(0);
    int t = 1, kase = 0; // cin >> t;
    while (t--) {
        // cout << "Case #" << ++kase << ":";
        marmot0814();
    }
}