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 = 1000000007;
long long N;
int K;
long long mat[45][45];
vector<vector<long long> > multiply(vector<vector<long long> > a,
                                    vector<vector<long long> > b) {
  vector<vector<long long> > c(a.size(), vector<long long>(b[0].size()));
  for (int i = 0; i < a.size(); i++)
    for (int j = 0; j < b[0].size(); j++)
      for (int k = 0; k < b.size(); k++) {
        c[i][j] += (a[i][k] * b[k][j]) % MOD;
        c[i][j] %= MOD;
      }
  return c;
}
vector<vector<long long> > power(vector<vector<long long> > a, long long p) {
  if (p == 1) return a;
  if (p % 2 == 1) return multiply(a, power(a, p - 1));
  vector<vector<long long> > b = power(a, p / 2);
  return multiply(b, b);
}
int main() {
  cin >> N >> K;
  for (int i = 0; i <= K; i++) {
    mat[i][0] = 1;
    for (int j = 1; j <= i; j++)
      mat[i][j] = (mat[i - 1][j] + mat[i - 1][j - 1]) % MOD;
  }
  vector<vector<long long> > T(2 * K + 3, vector<long long>(2 * K + 3));
  for (int i = 0; i <= K; i++)
    for (int j = 0; j <= i; j++) {
      T[i][K + 1 + j] = mat[i][j];
      T[K + 1 + i][j] = T[K + 1 + i][K + 1 + j] = mat[i][j];
    }
  T[2 * K + 2][2 * K + 1] = T[2 * K + 2][2 * K + 2] = 1;
  T = power(T, N);
  long long ans = 0;
  for (int i = 0; i < 2 * K + 2; i++) ans = (ans + T[2 * K + 2][i]) % MOD;
  cout << ans << endl;
}