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 long long mod = 1000000007;
int n, k;
long long ans, s;
struct matrix {
  long long mp[50 * 2][50 * 2];
} base, m;
int c[50][50];
matrix mul(matrix a, matrix b) {
  matrix temp;
  for (int i = 0; i <= n; i++) {
    for (int j = 0; j <= n; j++) {
      temp.mp[i][j] = 0;
      for (int k = 0; k <= n; k++) {
        temp.mp[i][j] = (temp.mp[i][j] + a.mp[i][k] * b.mp[k][j] % mod) % mod;
      }
      temp.mp[i][j] %= mod;
    }
  }
  return temp;
}
void pow(long long x) {
  while (x) {
    if (x & 1) {
      m = mul(m, base);
    }
    x >>= 1;
    base = mul(base, base);
  }
}
void init() {
  memset(c, 0, sizeof(c));
  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]) % mod;
    }
  }
  for (int i = 0; i <= k; i++) {
    for (int j = 0; j <= i; j++) {
      base.mp[i][j] = base.mp[i][j + k + 1] = base.mp[i + k + 1][j] =
          c[i][j] * 1LL;
    }
  }
}
int main() {
  init();
  ans = 0;
  base.mp[n][k] = base.mp[n][n] = 1;
  for (int i = 0; i <= n; i++) {
    m.mp[i][i] = 1;
  }
  pow(s);
  for (int i = 0; i < n; i++) {
    ans += m.mp[n][i];
  }
  cout << ans % mod << endl;
  return 0;
}