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;
int K = 90;
const int MOD = 1000000007;
vector<vector<long long> > mul(vector<vector<long long> > A,
                               vector<vector<long long> > B) {
  vector<vector<long long> > C(K + 1, vector<long long>(K + 1));
  for (int i = 0; i < K; i++)
    for (int j = 0; j < K; j++)
      for (int k = 0; k < K; k++)
        C[i][j] = (C[i][j] + (A[i][k] * B[k][j])) % MOD;
  return C;
}
vector<vector<long long> > pow(vector<vector<long long> > A, long long p) {
  if (p == 1) return A;
  if (p % 2) return mul(A, pow(A, p - 1));
  vector<vector<long long> > X = pow(A, p / 2);
  return mul(X, X);
}
int ways[100][100];
int main() {
  for (int i = 0; i < K; i++) {
    ways[i][0] = 1;
    for (int j = 1; j < i + 1; j++)
      ways[i][j] = (ways[i - 1][j] + ways[i - 1][j - 1]) % MOD;
  }
  long long n, k;
  cin >> n >> k;
  K = 2 * k + 3;
  vector<vector<long long> > G(K + 1, vector<long long>(K + 1));
  for (int i = 0; i < k + 1; i++)
    for (int j = i; j < k + 1; j++)
      G[k + 1 + i][j] = G[i][k + 1 + j] = G[k + 1 + i][k + 1 + j] = ways[j][i];
  G[2 * k + 2][2 * k + 2] = G[2 * k + 1][2 * k + 2] = 1;
  G = pow(G, n);
  long long ans = 0;
  for (int i = 0; i < 2 * k + 2; i++) ans = (ans + G[i][2 * k + 2]) % MOD;
  printf("%lld\n", ans);
  return 0;
}