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 long long MOD = 1000000000 + 7;
const int NICO = 100;
const int MAKI = 100;
struct M {
  long long a[NICO][NICO];
  M() { memset(a, 0, sizeof(a)); }
};
int n, k;
long long s, c[NICO][NICO];
void init() {
  n = 2 * k + 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];
    }
  }
}
M mul(M A, M B) {
  M 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] += A.a[i][t] * B.a[t][j];
        ans.a[i][j] %= MOD;
      }
    }
  }
  return ans;
}
M qpow(M A, long long t) {
  M ans;
  for (int i = 0; i < n; i++) ans.a[i][i] = 1;
  while (t) {
    if (t & 1) ans = mul(ans, A);
    A = mul(A, A);
    t /= 2;
  }
  return ans;
}
int main() {
  cin >> s >> k, init();
  M A;
  for (int i = 0; i <= k; ++i) {
    for (int j = 0; j <= i; ++j) {
      A.a[i][j] = A.a[i + k + 1][j] = A.a[i][j + k + 1] = c[i][j];
    }
  }
  A.a[2 * k + 2][k] = A.a[2 * k + 2][2 * k + 2] = 1;
  A = qpow(A, s);
  long long res = 0;
  for (int i = 0; i < 2 * k + 2; i++) {
    res = (res + A.a[2 * k + 2][i]) % MOD;
  }
  cout << res << endl;
}