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 Java8 (OpenJDK 1.8.0) (judge verdict: Accepted)


public class Main {


  private static void solve() {
    int n = ni();
    long k = nl();
    int[][] a = ntable(n, n);
    int mod = 1000000000 + 7;
    a = pow(a, k, mod);
    
    long ret = 0;
    for (int i = 0; i < n; i ++) {
      for (int j = 0; j < n; j ++) {
        ret += a[i][j];
        ret %= mod;
      }
    }
    System.out.println(ret);
  }
  
  public static int[][] pow(int[][] A, long e, int mod)
  {
      int[][] MUL = A;
      int n = A.length;
      int[][] C = new int[n][n];
      for(int i = 0;i < n;i++)C[i][i] = 1;
      for(;e > 0;e>>>=1) {
          if((e&1)==1)C = mul(C, MUL, mod);
          MUL = p2(MUL, mod);
      }
      return C;
  }
  
  // int行列の積
  // O(N^3)
  public static int[][] mul(int[][] A, int[][] B, int mod)
  {
      assert A[0].length == B.length;
      int m = A.length;
      int n = A[0].length;
      int o = B[0].length;
      int[][] C = new int[m][o];
      for(int i = 0;i < m;i++){
          for(int j = 0;j < o;j++){
              long sum = 0;
              for(int k = 0;k < n;k++){
                  sum += (long)A[i][k] * B[k][j];
                  sum %= mod;
              }
              C[i][j] = (int)sum;
          }
      }
      return C;
  }
  
  public static int[][] p2(int[][] A, int mod)
  {
      int n = A.length;
      int[][] C = new int[n][n];
      for(int i = 0;i < n;i++){
          for(int j = 0;j < n;j++){
              long sum = 0;
              for(int k = 0;k < n;k++){
                  sum += (long)A[i][k] * A[k][j];
                  sum %= mod;
              }
              C[i][j] = (int)sum;
          }
      }
      return C;
  }

  public static void main(String[] args) {
    new Thread(null, new Runnable() {
      @Override
      public void run() {
        long start = System.currentTimeMillis();
        String debug = args.length > 0 ? args[0] : null;
        if (debug != null) {
          try {
            is = java.nio.file.Files.newInputStream(java.nio.file.Paths.get(debug));
          } catch (Exception e) {
            throw new RuntimeException(e);
          }
        }
        reader = new java.io.BufferedReader(new java.io.InputStreamReader(is), 32768);
        solve();
        out.flush();
        tr((System.currentTimeMillis() - start) + "ms");
      }
    }, "", 64000000).start();
  }

  private static java.io.InputStream is = System.in;
  private static java.io.PrintWriter out = new java.io.PrintWriter(System.out);
  private static java.util.StringTokenizer tokenizer = null;
  private static java.io.BufferedReader reader;

  public static String next() {
    while (tokenizer == null || !tokenizer.hasMoreTokens()) {
      try {
        tokenizer = new java.util.StringTokenizer(reader.readLine());
      } catch (Exception e) {
        throw new RuntimeException(e);
      }
    }
    return tokenizer.nextToken();
  }

  private static double nd() {
    return Double.parseDouble(next());
  }

  private static long nl() {
    return Long.parseLong(next());
  }

  private static int[] na(int n) {
    int[] a = new int[n];
    for (int i = 0; i < n; i++)
      a[i] = ni();
    return a;
  }

  private static char[] ns() {
    return next().toCharArray();
  }

  private static long[] nal(int n) {
    long[] a = new long[n];
    for (int i = 0; i < n; i++)
      a[i] = nl();
    return a;
  }

  private static int[][] ntable(int n, int m) {
    int[][] table = new int[n][m];
    for (int i = 0; i < n; i++) {
      for (int j = 0; j < m; j++) {
        table[i][j] = ni();
      }
    }
    return table;
  }

  private static int[][] nlist(int n, int m) {
    int[][] table = new int[m][n];
    for (int i = 0; i < n; i++) {
      for (int j = 0; j < m; j++) {
        table[j][i] = ni();
      }
    }
    return table;
  }

  private static int ni() {
    return Integer.parseInt(next());
  }

  private static void tr(Object... o) {
    if (is != System.in)
      System.out.println(java.util.Arrays.deepToString(o));
  }
}