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++14 (GCC 5.4.1) (judge verdict: Accepted)

// #pragma GCC optimize("O3")
#include <bits/stdc++.h>

using namespace std;

using ll = long long;
using vll = vector<ll>;
using vii = vector<int>;
using pii = pair<int, int>;

const ll INF = numeric_limits<ll>::max(),
         MOD = 1e9 + 7;

const int INF_i = numeric_limits<int>::max(),
          MAX = 1e6;

using vvl = vector<vll>;

struct Matrix {
    vvl mat;

    Matrix(int n, int m, int e = 0) : mat(vvl(n, vll(m, e))) {}
    Matrix(int n, bool eye = false) : Matrix(n, n) {
        if (eye) {
            for (int i = 0; i < n; i++) {
                mat[i][i] = 1;
            }
        }
    }
    Matrix(vvl mat) : mat(mat) {}

    int rows() const { return mat.size(); }
    int cols() const { return mat[0].size(); }

    void operator*=(Matrix& rhs) {
        const int n = rows(), m = rhs.cols();
        vvl prod(n, vll(m, 0));
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < m; j++) {
                for (int k = 0; k < rhs.rows(); k++) {
                    prod[i][j] += mat[i][k] * rhs[k][j];
                    prod[i][j] %= MOD;
                }
            }
        }
        swap(mat, prod);
    }

    friend Matrix operator*(Matrix lhs, Matrix rhs) {
        lhs *= rhs;
        return lhs;
    }

    friend Matrix operator^(Matrix a, ll n) {
        Matrix res(a.cols(), true);
        for (; n; a *= a, n >>= 1)
            if (n & 1)
                res *= a;
        return res;
    }

    vll& operator[](int i) { return mat[i]; }

    friend ostream& operator<<(ostream& out, Matrix m) {
        out << '\n';
        for (auto v : m.mat) {
            for (auto e : v)
                out << e << ' ';
            out << '\n';
        }
        return out;
    }
};

int main() {
    cin.tie(NULL);
    ios_base::sync_with_stdio(false);
    // freopen("in.in", "r", stdin);

    int n;
    cin >> n;
    ll k;
    cin >> k;

    Matrix A(n), B(n, 1, 1);
    for (int i = 0; i < n; i++) {
        for (int j = 0; j < n; j++) {
            cin >> A[j][i];
        }
    }
    B = (A ^ k) * B;
    ll res = 0;
    for (int i = 0; i < n; i++) {
        (res += B[i][0]) %= MOD;
    }
    cout << res << '\n';
}