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 A[84][84], C[41][41], ans[84][84], tp[84][84];
void gen(int k) {
  for (int i = 0; i <= k; ++i) {
    C[i][0] = C[i][i] = 1;
  }
  for (int i = 2; i <= k; ++i) {
    for (int j = 1; j < i; ++j) {
      C[i][j] = (C[i - 1][j - 1] + C[i - 1][j]) % 1000000007;
    }
  }
  for (int i = 0; i <= k; ++i) {
    for (int j = i; j <= k; ++j) {
      A[i + k + 1][j] = A[i][j + k + 1] = A[i + k + 1][j + k + 1] = C[j][i];
    }
  }
  A[2 * k + 1][2 * k + 2] = 1;
  A[2 * k + 2][2 * k + 2] = 1;
  for (int i = 0; i < 2 * k + 3; ++i) {
    ans[i][i] = 1;
  }
}
void multi(long long C[][84], long long A[][84], long long B[][84], int n) {
  memset(tp, 0, sizeof(tp));
  for (int i = 0; i < n; ++i) {
    for (int j = 0; j < n; ++j) {
      for (int k = 0; k < n; ++k) {
        tp[i][j] = (tp[i][j] + A[i][k] * B[k][j] % 1000000007) % 1000000007;
      }
    }
  }
  memcpy(C, tp, sizeof(tp));
}
long long solve(long long n, int k) {
  gen(k);
  while (n) {
    if (n % 2) {
      multi(ans, A, ans, 2 * k + 3);
    }
    multi(A, A, A, 2 * k + 3);
    n /= 2;
  }
  long long res = 0;
  for (int i = 0; i < 2 * k + 2; ++i) {
    res = (res + ans[i][2 * k + 2]) % 1000000007;
  }
  return res;
}
int main() {
  long long n;
  int k;
  cin >> n >> k;
  cout << solve(n, k) << endl;
  return 0;
}