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: Wrong Answer)

    #include <iostream>
    #include <cstring>
    #include <vector>
    #include <set>
    #include <map>
    #include <sstream>
    #include <cstdio>
    #include <algorithm>
    #include <stack>
    #include <queue>
    #include <cmath>
    #include <iomanip>
    #include <fstream>
    //#include <bits/stdc++.h>
    using namespace std;
    using ll = long long;
    const ll INF = (1 << 30);
    const ll inf = (1LL << 60LL);
    const ll  maxn = 105;
    ll n;
    ll k;
    const ll MOD = 1e9 + 7;
    vector<vector<ll> > mat;
    struct matrix{
        vector<vector<ll> > v;
        matrix(){
            v.resize(n);
            for(ll i = 0; i < n; i ++){
                v[i].resize(n);
                for(ll j =0 ; j < n; j ++){
                    v[i][j] = 0;
                }
            }
        }
        matrix operator* (const matrix& tmp) const {
            matrix ret = matrix();
            for(ll i = 0; i < n; i ++){
                for(ll j = 0; j < n; j ++){
                    for(ll k = 0; k < n; k ++){
                        ret.v[i][k] += ((v[i][j] * tmp.v[j][k]) % MOD) % MOD;
                    }
                }
            }
            return ret;
        };
    };
    int main(int argc, const char * argv[]) {
        ios_base::sync_with_stdio(false);
        cin >> n >> k;
        mat.resize(n);
        for(ll i = 0; i < n; i ++){
            mat[i].resize(n);
        }
        for(ll i = 0; i < n; i ++){
            for(ll j = 0; j < n; j ++){
                cin >> mat[i][j];
            }
        }
        matrix ret = matrix();
        for(ll i = 0; i < n; i ++){
            ret.v[i][i] = 1;
        }
        matrix b = matrix();
        b.v = mat;
        while(k){
            if(k % 2 == 1){
                ret = (ret * b);
            }
            b = (b * b);
            k /= 2;
        }
        ll ans = 0;
        for(ll i =0 ; i < n; i ++){
            for(ll j = 0; j < n; j ++){
                ans += ret.v[i][j];
                ans %= MOD;
            }
        }
        printf("%lld\n", ans);
        return 0;
    }