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 (GCC 5.4.1) (judge verdict: Accepted)

#include<stdio.h>
        #define int long long
        #define fo(i,l,u)   for(i=l;i<u;i++)
        #define MOD         1000000007





        //int n;
    /*    void multiply(int n,int arr[][n] , int ary[][n]){
            int mat[n][n],i,j,k;
            fo(i,0,n){
                fo(j,0,n){
                    mat[i][j]=0;
                    fo(k,0,n){
                        mat[i][j]=(mat[i][j]+arr[i][k]*ary[k][j])%MOD;
                    }
                }
            }
            fo(i,0,n){
                fo(j,0,n){
                    //cout<<mat[i][j]<<" ";
                    arr[i][j]=mat[i][j];
                }
            }
            //cout<<endl;
        }

     */

        signed main()
        { int i,j,x,d,k,n,K;
      scanf("%ld%ld",&n,&K);
            int arr[n][n];
            fo(i,0,n){
                fo(j,0,n){
                scanf("%ld",&x);
                arr[i][j]=x;
                }
            }
            //cout<<K<<endl;
            int ans[n][n];
            fo(i,0,n){
                fo(j,0,n){
                    ans[i][j]=0;
                }
            }
            fo(i,0,n){
                ans[i][i]=1;
            }





            while(K>0){

              /*  fo(i,0,n){
                            fo(j,0,n){
                                cout<<ans[i][j]<<" ";
                                //arr[i][j]=mat[i][j];
                            }
                            cout<<endl;
                        }*/



                if(K%2==1){
                    //multiply(n,ans,arr);


                    int mat[n][n];
                        fo(i,0,n){
                            fo(j,0,n){
                                mat[i][j]=0;
                                fo(k,0,n){
                                    mat[i][j]=(mat[i][j]+ans[i][k]*arr[k][j])%MOD;
                                    //cout<<mat[i][]
                                }
                               // cout<<mat[i][j]<<" ";
                            }
                            //cout<<endl;
                        }
                        fo(i,0,n){
                            fo(j,0,n){
                                //cout<<mat[i][j]<<" ";
                                ans[i][j]=mat[i][j];
                            }
                        }


                }

                // cout<<arr[0][0]<<endl;

                int mat[n][n];

                        fo(i,0,n){
                            fo(j,0,n){
                                mat[i][j]=0;
                                fo(k,0,n){
                                    mat[i][j]=(mat[i][j]+arr[i][k]*arr[k][j])%MOD;
                                }
                            }
                        }
                        fo(i,0,n){
                            fo(j,0,n){
                                //cout<<mat[i][j]<<" ";
                                arr[i][j]=mat[i][j];
                            }
                        }



             //   multiply(n,arr,arr);





                K/=2;
            }
            x=0;
            fo(i,0,n){
                fo(j,0,n){
                    x=(x+ans[i][j])%MOD;
                }
            }
            printf("%ld",x);

            return 0;
        }