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;
const int M = 50;
const long long mod = 1e9 + 7;
struct mat {
  long long a[N][N];
  mat() { memset(a, 0, sizeof(a)); }
};
int n, k;
long long s, c[M][M];
void init() {
  cin >> s >> k;
  n = k * 2 + 3;
  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];
  }
}
mat mul(mat p, mat q) {
  mat ans;
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < n; j++) {
      for (int t = 0; t < n; t++) {
        ans.a[i][j] += p.a[i][t] * q.a[t][j];
        ans.a[i][j] %= mod;
      }
    }
  }
  return ans;
}
mat Qpow(mat base, long long t) {
  mat x;
  for (int i = 0; i < n; i++) x.a[i][i] = 1;
  while (t) {
    if (t & 1) x = mul(x, base);
    base = mul(base, base);
    t /= 2;
  }
  return x;
}
int main() {
  init();
  mat base;
  for (int i = 0; i <= k; i++) {
    for (int j = 0; j <= i; j++) {
      base.a[i][j] = base.a[i + k + 1][j] = base.a[i][j + k + 1] = c[i][j];
    }
  }
  base.a[2 * k + 2][k] = base.a[2 * k + 2][2 * k + 2] = 1;
  mat ans = Qpow(base, s);
  long long g = 0;
  for (int i = 0; i < 2 * k + 2; i++) {
    g = (g + ans.a[2 * k + 2][i]) % mod;
  }
  cout << g << endl;
  return 0;
}