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>
#pragma GCC optimize("O3")
#pragma GCC optimize("unroll-loops")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
using namespace std;
const int N = 500005;
const int INF = 1000000000;
long long mod = 1e9 + 7;
struct Matrix : vector<vector<long long> > {
  Matrix() {}
  Matrix(int n) { assign(n, vector<long long>(n, 0LL)); }
  void operator*=(Matrix s) {
    int n = size();
    Matrix ans(n), &x = *this;
    for (int i = 0; i < n; i++)
      for (int k = 0; k < n; k++)
        if (x[i][k])
          for (int j = 0; j < n; j++)
            ans[i][j] += x[i][k] * s[k][j], ans[i][j] %= mod;
    x = ans;
  }
};
Matrix Pow(Matrix x, long long m) {
  Matrix ans(x.size());
  for (int i = 0; i < x.size(); i++)
    for (int j = 0; j < x.size(); j++) ans[i][j] = int(i == j ? 1 : 0);
  while (m) {
    if (m & 1) ans *= x;
    x *= x;
    m >>= 1;
  }
  return ans;
}
int C[41][41];
int main() {
  long long n, k;
  cin >> n >> k;
  Matrix A(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]) % mod;
    for (int j = 0; j <= i; j++)
      A[i][j] = A[i][j + k + 1] = A[i + k + 1][j] = C[i][j];
  }
  A[2 * k + 2] = A[k];
  A[2 * k + 2][2 * k + 2] = 1;
  vector<int> X;
  for (int i = 0; i < k + 1; i++) X.push_back(1);
  for (int i = 0; i < k + 1; i++) X.push_back(1);
  X.push_back(1);
  Matrix B = Pow(A, n - 1);
  long long ans = 1;
  for (int i = 0; i < 2 * k + 2; i++)
    ans = (ans + B[2 * k + 2][i] * X[i]) % mod;
  cout << ans << endl;
}