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;
int n, k;
long long N;
int po[85];
int init[85][85], tmp[85][85], ans[85][85], c[85][85];
void MUL(int a[85][85]) {
  memset(tmp, 0, sizeof(tmp));
  for (int i = 0; i <= n; i++)
    for (int j = 0; j <= n; j++)
      for (int k = 0; k <= n; k++)
        if (a[k][j])
          tmp[i][j] = (tmp[i][j] + (long long)ans[i][k] * a[k][j]) % 1000000007;
  memcpy(ans, tmp, sizeof(tmp));
}
void POW(long long p) {
  if (p == 1) {
    memcpy(ans, init, sizeof(ans));
    return;
  }
  POW(p / 2);
  MUL(ans);
  if (p & 1) MUL(init);
}
int main() {
  po[0] = 1;
  for (int i = 1; i <= 45; i++) po[i] = po[i - 1] * 2 % 1000000007;
  c[0][0] = 1;
  for (int i = 1; i <= 45; i++) {
    c[i][0] = 1;
    for (int j = 1; j <= i; j++)
      c[i][j] = (c[i - 1][j - 1] + c[i - 1][j]) % 1000000007;
  }
  while (scanf("%I64d%d", &N, &k) != -1) {
    memset(init, 0, sizeof(init));
    for (int i = 0; i <= k; i++)
      for (int j = 0; j <= i; j++) init[j][i] = c[i][j];
    for (int i = 0; i <= k; i++)
      for (int j = 0; j <= i; j++)
        init[j + k + 1][i] = (long long)c[i][j] * po[i - j] % 1000000007;
    for (int i = 0; i <= k; i++) init[i][i + k + 1] = 1;
    n = 2 * k + 2;
    init[k][n] = init[n][n] = 1;
    POW(N + 1);
    printf("%d\n", ans[0][n]);
  }
  return 0;
}