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 Python (3.4.3) (judge verdict: Runtime Error)

#include <iostream>
#include <algorithm>
#include <map>
#include <set>
#include <queue>
#include <string>
 
using namespace std;
 
#define rip(i, n, s) for (int i = (s);i < ( int )( n ); i++)
#define all(a) (a).begin(), (a).end()
typedef long long ll; 

struct modint {
	long long num;
	long long p;
	modint() {
		num = 0;
		p = 1000000007;
	}
	modint(int x) {
		p = 1000000007;
		num = (long long)x%p;
	}
	modint(long long x) {
		p = 1000000007;
		num = x % p;
	}
	modint operator+(const modint &other) {
		modint ret;
		ret.p = p;
		ret.num = (num + other.num) % p;
		return ret;
	}
	modint operator+(const long long &other) {
		modint ret;
		ret.p = p;
		ret.num = (num + (other%p)) % p;
		return ret;
	}
	modint operator-(const modint &other) {
		modint ret;
		ret.p = p;
		ret.num = (num - other.num + p) % p;
		return ret;
	}
	modint operator-(const long long &other) {
		modint ret;
		ret.p = p;
		ret.num = (num - (other%p) + p) % p;
		return ret;
	}
	modint operator*(const modint &other) {
		modint ret;
		ret.p = p;
		ret.num = (num*other.num) % p;
		return ret;
	}
	modint operator*(const long long &other) {
		modint ret;
		ret.p = p;
		ret.num = (num*(other%p)) % p;
		return ret;
	}
	void operator+=(const long long &other) {
		num = (num + other) % p;
	}
	void operator+=(const modint &other) {
		num = (num + other.num) % p;
	}
	void operator-=(const long long &other) {
		num = (num - other + p) % p;
	}
	void operator-=(const modint &other) {
		num = (num - other.num + p) % p;
	}
	void operator*=(const long long &other) {
		num = (num*other) % p;
	}
	void operator*=(const modint &other) {
		num = (num*other.num) % p;
	}
};

template<typename T>
struct matrix{
	int size;
	vector<vector<T> > table;
	matrix(int n){
		size=n;
		table=vector<vector<T> >(n,vector<T>(n));
	}
	matrix<T> operator*(const matrix<T> &othor){
		matrix<T> ret(size);
		for(int i=0;i<size;i++){
			for(int j=0;j<size;j++){
				for(int k=0;k<size;k++){
					ret.table[i][j]+=table[i][k]*othor.table[k][j];
				}
			}
		}
		return ret;
	}
	T all_sum(){
		T ret = table[0][0];
		for(int i=1;i<size;i++){
			ret+=table[0][i];
		}
		for(int i=1;i<size;i++){
			for(int j=0;j<size;j++){
				ret+=table[i][j];
			}
		}
		return ret;
	}
};

template<typename T>
T pow(ll n,T &da){
	if (n==1){
		return da;
	}
	if (n%2){
		return pow(n-1,da)*da;
	}
	else{
		T ret = pow(n/2,da);
		return ret*ret;
	}
}

int main(){
	int n;
	ll k;
	cin >> n >> k;
	matrix<modint> mt(n);
	rip(i,n,0){
		rip(j,n,0){
			cin >> mt.table[i][j].num;
		}
	}
	mt=pow(k,mt);
	cout << mt.all_sum().num << endl;
}