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 N = 111, P = 1e9 + 7;
int k;
long long n;
long long C[N][N], pw[N];
void upd(long long &a, long long b) {
  a += b % P;
  if (a >= P) a -= P;
}
void init() {
  pw[0] = 1;
  for (int i = (1); i < (N); i++) pw[i] = pw[i - 1] * 2 % P;
  for (int i = (0); i < (N); i++) C[i][0] = C[i][i] = 1;
  for (int i = (0); i < (N); i++)
    for (int j = (1); j < (i); j++)
      C[i][j] = (C[i - 1][j] + C[i - 1][j - 1]) % P;
}
struct M {
  long long a[N][N];
  M() { memset(a, 0, sizeof(a)); }
  M operator*(const M &c) const {
    M res;
    int n = 2 * k + 3;
    for (int i = (0); i < (n); i++)
      for (int j = (0); j < (n); j++)
        for (int k = (0); k < (n); k++)
          upd(res.a[i][j], a[i][k] * c.a[k][j] % P);
    return res;
  }
};
M kpow(M a, long long b) {
  M res;
  for (int i = (0); i < (2 * k + 3); i++) res.a[i][i] = 1;
  while (b) {
    if (b & 1) res = res * a;
    a = a * a;
    b >>= 1;
  }
  return res;
}
int main() {
  init();
  while (~scanf("%lld%d", &n, &k)) {
    if (n == 1) {
      puts("1");
      continue;
    }
    M a;
    for (int i = (0); i < (k + 1); i++) {
      for (int j = (i); j < (k + 1); j++) a.a[i][j] = C[k - i][k - j];
      for (int j = (k + 1 + i); j < (2 * k + 2); j++)
        a.a[i][j] = C[k - i][k - j + k + 1] * pw[j - i - k - 1] % P;
    }
    for (int i = (k + 1); i < (2 * k + 2); i++) a.a[i][i - k - 1] = 1;
    a.a[2 * k + 2][0] = a.a[2 * k + 2][2 * k + 2] = 1;
    a = kpow(a, n - 2);
    long long ans = 0;
    for (int i = (0); i < (k + 1); i++)
      upd(ans, (a.a[0][i] + a.a[2 * k + 2][i]) * pw[k - i + 1] % P);
    for (int i = (k + 1); i < (2 * k + 2); i++)
      upd(ans, a.a[0][i] + a.a[2 * k + 2][i]);
    upd(ans, a.a[0][2 * k + 2] + a.a[2 * k + 2][2 * k + 2]);
    printf("%lld\n", ans);
  }
  return 0;
}