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 unsigned long long mod = 1e9 + 7;
unsigned long long n, C[45][45];
int k;
void init() {
  for (int i = 0; i < 44; i++) {
    C[i][0] = C[i][i] = 1;
    if (i == 0) continue;
    for (int j = 1; j <= i; j++)
      C[i][j] = (C[i - 1][j] % mod + C[i - 1][j - 1] % mod) % mod;
  }
}
vector<vector<unsigned long long> > mul(
    vector<vector<unsigned long long> > &A,
    vector<vector<unsigned long long> > &B) {
  vector<vector<unsigned long long> > C(
      A.size(), vector<unsigned long long>(B[0].size()));
  for (int i = 0; i < A.size(); ++i) {
    for (int k = 0; k < B.size(); ++k) {
      for (int j = 0; j < B[0].size(); ++j)
        C[i][j] = (C[i][j] + A[i][k] % mod * B[k][j] % mod) % mod;
    }
  }
  return C;
}
vector<vector<unsigned long long> > qpow(vector<vector<unsigned long long> > A,
                                         unsigned long long n) {
  vector<vector<unsigned long long> > B(A.size(),
                                        vector<unsigned long long>(A.size()));
  for (int i = 0; i < A.size(); ++i) B[i][i] = 1;
  while (n > 0) {
    if (n & 1) B = mul(B, A);
    A = mul(A, A);
    n >>= 1;
  }
  return B;
}
int main() {
  init();
  scanf("%lld%d", &n, &k);
  int len = 2 * k + 3;
  vector<vector<unsigned long long> > a(len, vector<unsigned long long>(len));
  a[0][0] = 1;
  for (int i = 1; i <= k + 1; ++i) a[0][i] = a[0][i + k + 1] = C[k][k + 1 - i];
  for (int i = 1; i <= k + 1; ++i)
    for (int j = i; j <= k + 1; ++j)
      a[i][j] = a[i][j + k + 1] = C[k + 1 - i][k + 1 - j];
  for (int i = k + 2; i < len; ++i)
    for (int j = i - k - 1; j <= k + 1; ++j)
      a[i][j] = C[k + k + 2 - i][k + 1 - j];
  a = qpow(a, n - 1);
  unsigned long long ans = 0;
  for (int i = 0; i < len; ++i) ans = (ans + a[0][i]) % mod;
  printf("%lld\n", ans);
  return 0;
}