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 = 45;
const long long mod = 1e9 + 7;
int k, n, c[N][N];
struct matrix {
  long long a[N * 2][N * 2];
};
long long s, ans;
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])) % mod;
      ans.a[i][j] %= mod;
    }
  return ans;
}
void Pow(long long k) {
  while (k) {
    if (k & 1) m = mul(m, base);
    base = mul(base, base);
    k >>= 1;
  }
}
int main() {
  scanf("%lld%d", &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 - 1); ++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)
      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 - 1); ++i) (ans += m.a[n][i]) %= mod;
  return 0 * printf("%lld\n", ans);
}