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>
using namespace std;
const int mod = 1000000007;
struct mat {
  long long a[90][90];
};
mat mul(mat A, mat B, int n) {
  mat C;
  int i, j, k;
  for (i = 0; i < n; i++)
    for (j = 0; j < n; j++) {
      C.a[i][j] = 0;
      for (k = 0; k < n; k++) {
        C.a[i][j] = (C.a[i][j] + A.a[i][k] * B.a[k][j]) % mod;
      }
    }
  return C;
}
long long C[50][50];
void init() {
  int i, j;
  for (i = 0; i < 50; i++) C[i][0] = 1;
  for (i = 1; i < 50; i++)
    for (j = 1; j <= i; j++) {
      C[i][j] = (C[i - 1][j] + C[i - 1][j - 1]) % mod;
    }
}
int main() {
  init();
  mat S, D;
  int k, i, j;
  long long m;
  cin >> m >> k;
  int n = 2 * k + 3;
  for (i = 0; i < n; i++)
    for (j = 0; j < n; j++) {
      S.a[i][j] = D.a[i][j] = 0;
      if (i == j) S.a[i][j] = 1;
    }
  for (i = 0; i <= k; i++) {
    int d = k + 1;
    for (j = 0; j <= i; j++) {
      D.a[i][j + d] = C[i][j];
    }
  }
  for (i = k + 1; i <= 2 * k + 1; i++) {
    int d = k + 1;
    for (j = 0; j <= i - k - 1; j++) {
      D.a[i][j + d] = D.a[i][j] = C[i - k - 1][j];
    }
  }
  D.a[2 * k + 2][2 * k + 1] = D.a[2 * k + 2][2 * k + 2] = 1;
  while (m) {
    if (m & 1) S = mul(S, D, n);
    m >>= 1;
    D = mul(D, D, n);
  }
  long long ans = 0;
  for (i = 0; i <= 2 * k + 1; i++) ans = (ans + S.a[2 * k + 2][i]) % mod;
  cout << ans << endl;
}