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 <bits/stdc++.h>

using namespace std;
typedef long long ll;

ll md = 1e9+7;
vector < vector < ll > > mul(vector < vector < ll > > v1, vector <vector < ll > > v2)
{
    vector < vector < ll >> v(v1.size() , vector < ll> (v1[0].size()));
    for (int i = 0; i < v1.size() ; ++i)
        for (int j = 0 ; j < v1[0].size() ; ++j)
        {
            v[i][j] = 0;
            for (int k = 0 ; k < v1[0].size() ; ++k)
                v[i][j] += (v1[i][k] * v2[k][j])%md,
                v[i][j] %= md;
        }
    return v;
}
vector < vector < ll > > po(vector < vector < ll > > b, ll p)
{
    if (p == 1) return b;
    if (p & 1) return mul(b, po(mul(b, b), p/2));
    return po(mul(b, b), p/2);
}

int main()
{
    int n;
    ll k;
    scanf("%d%lld", &n, &k);
    vector < vector < ll > > v (n, vector < ll>(n));
    for (int i = 0; i < n; ++i)
        for (int j = 0; j < n; ++j)
            scanf("%lld", &v[i][j]);
    auto it = po(v, k);
    ll ans = 0;
    for (auto u : it)
        for (auto v : u)
        {
            ans += v;
            if (ans >= md)
                ans -= md;
        }
    printf("%lld\n", ans);

    return 0;
}