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[100][100];
} base, ans;
long long n, k, N;
long long c[100][100];
void mul(mat &c, mat a, mat b) {
  for (int i = 0; i <= N; ++i)
    for (int j = 0; j <= N; ++j) c.a[i][j] = 0;
  for (int k = 0; k <= N; ++k)
    for (int i = 0; i <= N; ++i)
      for (int j = 0; j <= N; ++j)
        c.a[i][j] = (c.a[i][j] + a.a[i][k] * b.a[k][j] % mod) % mod;
}
int main() {
  cin >> n >> k;
  c[0][0] = 1;
  for (int i = 1; i <= k; ++i) {
    c[i][0] = c[i][i] = 1;
    for (int j = 1; j <= i - 1; ++j)
      c[i][j] = (c[i - 1][j] + c[i - 1][j - 1]) % mod;
  }
  N = 2 * k + 2;
  for (int i = 0; i <= N - 1; ++i) ans.a[0][i] = 1;
  base.a[N][N] = base.a[N - 1][N] = 1;
  for (int i = 0; i <= k; ++i)
    for (int j = 0; j <= i; ++j)
      base.a[k + 1 + j][i] = base.a[j][i + k + 1] =
          base.a[k + 1 + j][k + 1 + i] = c[i][j];
  while (n) {
    if (n & 1) mul(ans, ans, base);
    mul(base, base, base);
    n >>= 1;
  }
  cout << ans.a[0][N] << endl;
  return 0;
}