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;
long long t, k, n;
long long c[41][41];
struct mat {
  long long a[90][90];
  struct mat operator*(const struct mat& b) const {
    struct mat ans;
    int i, j, y;
    for (i = 0; i < n; i++)
      for (j = 0; j < n; j++) ans.a[i][j] = 0;
    for (i = 0; i < n; i++)
      for (j = 0; j < n; j++)
        for (y = 0; y < n; y++) {
          ans.a[i][j] += a[i][y] * b.a[y][j];
          ans.a[i][j] %= 1000000007;
        }
    return ans;
  }
};
struct mat qpow(struct mat a, long long asd) {
  int i, j;
  struct mat ans;
  for (i = 0; i < n; i++)
    for (j = 0; j < n; j++)
      if (i == j)
        ans.a[i][j] = 1;
      else
        ans.a[i][j] = 0;
  while (asd) {
    if (asd & 1) ans = ans * a;
    asd = asd >> 1;
    a = a * a;
  }
  return ans;
}
void initialize() {
  int i, j;
  cin >> t >> k;
  n = 2 * k + 3;
  for (i = 0; i <= k; i++) {
    c[i][0] = c[i][i] = 1;
    for (j = 1; j < i; j++) c[i][j] = c[i - 1][j - 1] + c[i - 1][j];
  }
}
int main() {
  long long ans = 0;
  int i, j;
  initialize();
  struct mat p;
  for (i = 0; i < n; i++)
    for (j = 0; j < n; j++) p.a[i][j] = 0;
  for (i = 0; i <= k; i++) {
    for (j = 0; j <= i; j++) {
      p.a[i][j] = c[i][j];
      p.a[i + 1 + k][j] = c[i][j];
      p.a[i][j + 1 + k] = c[i][j];
    }
  }
  p.a[2 * k + 2][k] = 1;
  p.a[2 * k + 2][2 * k + 2] = 1;
  p = qpow(p, t);
  for (i = 0; i < 2 * k + 2; i++) {
    ans += p.a[2 * k + 2][i];
    ans %= 1000000007;
  }
  cout << ans << endl;
}