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 <fstream>
#include <cmath>  
#include <cstdlib>
#include <ctime>
#include <algorithm>
#include <numeric>
#include <functional>
#include <string> 
#include <vector>
#include <bitset>
#include <map>
#include <set>
#include <stack>
#include <queue>
#include <deque>

using namespace std;
using ll = long long;

template<class T> using V = vector<T>;
template<class T, class U> using P = pair<T, U>;

#define REP(i,n) for(int i = 0; i < int(n); i++)
#define FOR(i, m, n) for(int i = int(m);i < int(n);i++)
#define ALL(obj) (obj).begin(),(obj).end()

const ll MOD = (ll)1e9 + 7;
const ll HINF = (ll)1e18;
const ll LINF = (ll)1e9;
const long double PI = 3.1415926535897932384626433;

template<class T> void corner(bool flg, T hoge) {if (flg) {cout << hoge << endl;exit(0);}else return;}
template <class T, class U>ostream &operator<<(ostream &o, const map<T, U>&obj) {o << "{"; for (auto &x : obj) o << " {" << x.first << " : " << x.second << "}" << ","; o << " }"; return o;}
template <class T>ostream &operator<<(ostream &o, const set<T>&obj) {o << "{"; for (auto itr = obj.begin(); itr != obj.end(); ++itr) o << (itr != obj.begin() ? ", " : "") << *itr; o << "}"; return o;}
template <class T>ostream &operator<<(ostream &o, const vector<T>&obj) {o << "{"; for (int i = 0; i < (int)obj.size(); ++i)o << (i > 0 ? ", " : "") << obj[i]; o << "}"; return o;}
template <class T, class U>ostream &operator<<(ostream &o, const pair<T, U>&obj) {o << "{" << obj.first << ", " << obj.second << "}"; return o;}
template <template <class tmp>  class T, class U> ostream &operator<<(ostream &o, const T<U> &obj) {o << "{"; for (auto itr = obj.begin(); itr != obj.end(); ++itr)o << (itr != obj.begin() ? ", " : "") << *itr; o << "}"; return o;}
void print(void) {cout << endl;}
template <class Head> void print(Head&& head) {cout << head;print();}
template <class Head, class... Tail> void print(Head&& head, Tail&&... tail) {cout << head << " ";print(forward<Tail>(tail)...);}
void YN(bool flg) {cout << ((flg) ? "YES" : "NO") << endl;}
void Yn(bool flg) {cout << ((flg) ? "Yes" : "No") << endl;}
void yn(bool flg) {cout << ((flg) ? "yes" : "no") << endl;}

//MatrixRepeatedMultiplicationMod
vector<vector<long long>> MatrixRepeatedMultiplicationMod(vector<vector<long long>> mat, long long K, long long mod) {
    int N = mat.size();
    vector<vector<long long>> res(N, vector<long long>(N)), tmp(N, vector<long long>(N));

    for (int i = 0; i < N; ++i) for (int j = 0; j < N; ++j) res[i][j] = (i == j);
    for (; K > 0; K /= 2) {
        if (K & 1) {
            for (int i = 0; i < N; ++i) for (int j = 0; j < N; ++j) tmp[i][j] = 0;
            for (int i = 0; i < N; ++i) for (int j = 0; j < N; ++j) for (int k = 0; k < N; ++k) (tmp[i][j] += (mat[i][k] * res[k][j]) % mod) %= mod;
            res = tmp;
        }
        for (int i = 0; i < N; ++i) for (int j = 0; j < N; ++j) tmp[i][j] = 0;
        for (int i = 0; i < N; ++i) for (int j = 0; j < N; ++j) for (int k = 0; k < N; ++k) (tmp[i][j] += (mat[i][k] * mat[k][j]) % mod) %= mod;
        mat = tmp;
    }
    return res;
}


int main() {
	ll N, K;
	cin >> N >> K;
	V<V<ll>> a(N, V<ll>(N));
	REP(i, N) REP(j, N) cin >> a[i][j];
	a = MatrixRepeatedMultiplicationMod(a, K, MOD);
	ll ans = 0;
	REP(i, N) REP(j, N) (ans += a[i][j]) %= MOD;
	cout << ans << endl;

	return 0;
}