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 N = 90, mod = 1e9 + 7;
long long n;
int m, k, ans, C[N][N];
struct mat {
  int a[N][N];
} R, T;
mat mul(mat a, mat b) {
  mat c;
  for (int i = 0; i <= m; i++)
    for (int j = 0; j <= m; j++) c.a[i][j] = 0;
  for (int i = 0; i <= m; i++)
    for (int j = 0; j <= m; j++)
      for (int k = 0; k <= m; k++)
        c.a[i][j] = (c.a[i][j] + 1ll * a.a[i][k] * b.a[k][j] % mod) % mod;
  return c;
}
int main() {
  scanf("%lld%d", &n, &k);
  m = k * 2 + 2;
  for (int i = 0; i <= k; i++) C[i][0] = 1;
  for (int i = 1; i <= k; i++)
    for (int j = 1; j <= i; j++)
      C[i][j] = (C[i - 1][j - 1] + C[i - 1][j]) % mod;
  for (int i = 0; i <= k; i++)
    for (int j = 0; j <= i; j++)
      T.a[i][j] = T.a[i][j + k + 1] = T.a[i + k + 1][j] = C[i][j];
  T.a[m][k] = T.a[m][m] = 1;
  for (int i = 0; i <= m; i++) R.a[i][i] = 1;
  for (; n; n >>= 1, T = mul(T, T))
    if (n & 1) R = mul(R, T);
  for (int i = 0; i < m; i++) (ans += R.a[m][i]) %= mod;
  printf("%d\n", ans);
  return 0;
}