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 P = 1e9 + 7;
long long s, c[60][60];
long long n, k, ans;
void Add(long long &x, long long y) {
  x += y;
  while (x >= P) x -= P;
}
struct Martix {
  long long a[90][90];
  Martix() { memset(a, 0, sizeof a); }
  Martix operator*(const Martix &B) const {
    Martix C;
    for (int k = 0; k < n; k++)
      for (int i = 0; i < n; i++)
        for (int j = 0; j < n; j++)
          Add(C.a[i][j], 1ll * a[i][k] * B.a[k][j] % P);
    return C;
  }
  Martix operator^(long long B) const {
    Martix A = (*this), C;
    for (int i = 0; i < n; i++) C.a[i][i] = 1;
    for (; B; B >>= 1, A = A * A)
      if (B & 1) C = C * A;
    return C;
  }
} E;
int main() {
  scanf("%lld%d", &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];
  }
  for (int i = 0; i <= k; i++)
    for (int j = 0; j <= i; j++)
      E.a[i][j] = E.a[i + k + 1][j] = E.a[i][j + k + 1] = c[i][j];
  E.a[2 * k + 2][k] = E.a[2 * k + 2][2 * k + 2] = 1;
  E = E ^ s;
  for (int i = 0; i < 2 * k + 2; i++) Add(ans, E.a[2 * k + 2][i]);
  printf("%d\n", ans);
  return 0;
}