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)

#include <iostream>
#include <vector>

using namespace std;
typedef long long ll;
typedef vector<vector<ll>> matrix;

constexpr ll MOD = 1e9 + 7;

int main() {
    ll n, k;
    cin >> n >> k;
    vector<matrix> doubling(70, matrix(n, vector<ll>(n)));
    for (int i = 0; i < n; i++) {
        for (int j = 0; j < n; j++) {
            cin >> doubling[0][i][j];
        }
    }

    for (int t = 1; t < 70; t++) {
        for (int from = 0; from < n; from++) {
            for (int mid = 0; mid < n; mid++) {
                for (int to = 0; to < n; to++) {
                    doubling[t][from][to] += doubling[t - 1][from][mid] * doubling[t - 1][mid][to];
                    doubling[t][from][to] %= MOD;
                }
            }
        }
    }

    vector<ll> ret(n, 1);
    for (int t = 0; t < 70; t++) {
        if (k & 1) {
            vector<ll> after(n);
            for (int from = 0; from < n; from++) {
                for (int to = 0; to < n; to++) {
                    after[to] += ret[from] * doubling[t][from][to];
                    after[to] %= MOD;
                }
            }
            ret = after;
        }
        k >>= 1;
    }

    ll ans = 0;
    for (auto x:ret)ans += x, ans %= MOD;
    cout << ans << endl;
    return 0;
}