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 maxn = 45;
long long s, ans;
int k, n, c[maxn][maxn];
struct matrix {
  long long a[maxn * 2][maxn * 2];
};
matrix base, m;
matrix mul(matrix x, matrix y) {
  matrix ans;
  for (int i = 0; i <= n; i++)
    for (int j = 0; j <= n; j++) {
      ans.a[i][j] = 0;
      for (int k = 0; k <= n; k++)
        ans.a[i][j] += ((x.a[i][k]) * (y.a[k][j])) % 1000000007LL;
      ans.a[i][j] %= 1000000007LL;
    }
  return ans;
}
void pow(long long k) {
  while (k) {
    if (k & 1) m = mul(m, base);
    base = mul(base, base);
    k >>= 1;
  }
}
int main(void) {
  cin >> s >> k;
  n = k * 2 + 2;
  for (int i = 0; i <= k; i++) {
    c[i][0] = c[i][i] = 1;
    for (int j = 1; j < i; j++)
      c[i][j] = (c[i - 1][j - 1] + c[i - 1][j]) % 1000000007LL;
  }
  for (int i = 0; i <= k; i++)
    for (int j = 0; j <= i; j++)
      base.a[i][j] = base.a[i][j + k + 1] = base.a[i + k + 1][j] = c[i][j];
  base.a[n][k] = base.a[n][n] = 1;
  for (int i = 0; i <= n; i++) m.a[i][i] = 1;
  pow(s);
  for (int i = 0; i < n; i++) ans += m.a[n][i];
  cout << ans % 1000000007LL << endl;
  return 0;
}