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)

#define _CRT_SECURE_NO_WARNINGS
#include "bits/stdc++.h"

#define PB push_back
#define PF push_front
#define LB lower_bound
#define UB upper_bound
#define fr(x) freopen(x,"r",stdin)
#define fw(x) freopen(x,"w",stdout)
#define iout(x) printf("%d\n",x)
#define lout(x) printf("%lld\n",x)
#define REP(x,l,u) for(ll x = l;x<u;x++)
#define RREP(x,l,u) for(ll x = l;x>=u;x--)
#define complete_unique(a) a.erase(unique(a.begin(),a.end()),a.end())
#define mst(x,a) memset(x,a,sizeof(x))
#define all(a) a.begin(),a.end()
#define PII pair<int,int>
#define PLL pair<ll,ll>
#define MP make_pair
#define sqr(x) ((x)*(x))
#define lowbit(x) ((x)&(-(x)))
#define lson (ind<<1)
#define rson (ind<<1|1)
#define se second
#define fi first
#define sz(x) ((int)x.size())
#define EX0 exit(0);

typedef  long long ll;
typedef unsigned long long ull;
typedef double db;
typedef long double ld;
using namespace std;
typedef vector<ll> VLL;
typedef vector<int> VI;
const int block_size = 320;
typedef complex<ll> point;
const ll mod = 1e9 + 7;
const ll inf = 1e9 + 7;
const ll INF = 1e15;
const ld eps = 1e-9;
const db PI = atan(1) * 4;


template<typename T, typename S>inline bool upmin(T&a, const S&b) { return a > b ? a = b, 1 : 0; }
template<typename T, typename S>inline bool upmax(T&a, const S&b) { return a < b ? a = b, 1 : 0; }

ll gcd(ll a, ll b) { return b == 0 ? a : gcd(b, a%b); }
ll pow(ll a, ll b) {
	if (b == 0) {
		return 1;
	}
	if (b == 1) {
		return a;
	}
	ll ans = pow(a, b / 2);
	ans *= ans;
	ans *= b % 2 == 1 ? a : 1;
	return ans;
}

vector<VLL> mult(vector<VLL> A, vector<VLL> B) {
	int n = sz(A);
	vector<VLL> C(n, VLL(n));
	REP(i, 0, n) {
		REP(j, 0, n) {
			REP(k, 0, n) {
				C[i][j] += A[i][k] * B[k][j];
				C[i][j] %= mod;
			}
		}
	}
	return C;
}
vector<VLL> matrixpow(vector<VLL> A, ll b) {
	if (b == 1) {
		return A;
	}
	vector<VLL> ans = matrixpow(A, b / 2);
	ans = mult(ans, ans);
	ans = b % 2 == 1 ? mult(ans, A) : ans;
	return ans;
}

namespace SOLVE {
	void main() {
		ll n, k;
		cin >> n >> k;
		
		vector<VLL> adjmatrix(n, VLL(n));
		REP(i, 0, n) {
			REP(j, 0, n) {
				cin >> adjmatrix[i][j];
			}
		}
		vector<VLL> ans = matrixpow(adjmatrix, k);
		ll count = 0;
		REP(i, 0, n) {
			REP(j, 0, n) {
				// cout << ans[i][j] << ' ';
				count += ans[i][j];
				count %= mod;
			} // cout << endl;
		}

		cout << count << endl;

	}
}


signed main() {
	ios_base::sync_with_stdio(false);
	cin.tie(NULL);

	int t;
	//    in(t);
	t = 1;
	while (t--) {
		SOLVE::main();

	}

	return 0;
}