Problem: Everyone knows what the Fibonacci sequence is. This sequence can be defined by the recurrence relation: 

F1 = 1, F2 = 2, Fi = Fi - 1 + Fi - 2 (i > 2).

We'll define a new number sequence Ai(k) by the formula: 

Ai(k) = Fi × ik (i ≥ 1).

In this problem, your task is to calculate the following sum: A1(k) + A2(k) + ... + An(k). The answer can be very large, so print it modulo 1000000007 (109 + 7).

Input

The first line contains two space-separated integers n, k (1 ≤ n ≤ 1017; 1 ≤ k ≤ 40).

Output

Print a single integer — the sum of the first n elements of the sequence Ai(k) modulo 1000000007 (109 + 7).

Examples

Input

1 1


Output

1


Input

4 1


Output

34


Input

5 2


Output

316


Input

7 4


Output

73825

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
int c[51][51];
long long N;
int K, i, j, ans;
int top, dd[105][105], f[105][105], tmp[105][105];
int power(long long b) {
  for (int i = 0; i <= top; i++)
    for (int j = 0; j <= top; j++) dd[i][j] = f[i][j];
  while (b) {
    if (b & 1) {
      for (int i = 0; i <= top; i++)
        for (int j = 0; j <= top; j++) tmp[i][j] = 0;
      for (int i = 0; i <= top; i++)
        for (int j = 0; j <= top; j++)
          for (int k = 0; k <= top; k++)
            tmp[i][j] =
                (tmp[i][j] + (long long)dd[i][k] * f[k][j]) % 1000000007;
      for (int i = 0; i <= top; i++)
        for (int j = 0; j <= top; j++) dd[i][j] = tmp[i][j];
    }
    b >>= 1;
    for (int i = 0; i <= top; i++)
      for (int j = 0; j <= top; j++) tmp[i][j] = 0;
    for (int i = 0; i <= top; i++)
      for (int j = 0; j <= top; j++)
        for (int k = 0; k <= top; k++)
          tmp[i][j] = (tmp[i][j] + (long long)f[i][k] * f[k][j]) % 1000000007;
    for (int i = 0; i <= top; i++)
      for (int j = 0; j <= top; j++) f[i][j] = tmp[i][j];
  }
  return dd[top - 1][top];
}
int main() {
  scanf("%I64d%d", &N, &K);
  c[0][0] = 1;
  for (i = 1; i <= K; i++) {
    c[i][0] = 1;
    for (j = 1; j <= i; j++)
      c[i][j] = (c[i - 1][j - 1] + c[i - 1][j]) % 1000000007;
  }
  for (i = 0; i <= 2 * K + 1; i++)
    for (j = 0; j <= 2 * K + 1; j++) {
      if (i <= K && j <= K) continue;
      if (i <= K && j > K) f[i][j] = c[2 * K + 1 - j][K - i];
      if (i > K && j <= K) f[i][j] = c[K - j][2 * K + 1 - i];
      if (i > K && j > K) f[i][j] = c[2 * K + 1 - j][2 * K + 1 - i];
    }
  f[2 * K + 2][2 * K + 2] = 1;
  f[K + 1][2 * K + 2] = 1;
  top = 2 * K + 2;
  ans = power(N);
  printf("%d\n", ans);
}