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 maxn = 90;
const int maxk = 50;
const int mod = 1000000007;
long long n, k, c[maxk][maxk];
void init() {
  for (int i = 0; i <= 40; i++) {
    c[i][0] = c[i][i] = 1;
    for (int j = 1; j < i; j++) c[i][j] = (c[i - 1][j] + c[i - 1][j - 1]) % mod;
  }
}
struct Matrix {
  int n, m;
  long long s[maxn][maxn];
  void init(int n, int m) {
    this->n = n;
    this->m = m;
    memset(s, 0, sizeof(s));
  }
  Matrix operator*(const Matrix& rhs) {
    Matrix ans;
    ans.init(n, rhs.m);
    for (int i = 0; i < n; i++)
      for (int j = 0; j < rhs.m; j++)
        for (int k = 0; k < m; k++)
          ans.s[i][j] = (ans.s[i][j] + s[i][k] * rhs.s[k][j]) % mod;
    return ans;
  }
} tmp, ans;
Matrix qpow(Matrix a, long long n) {
  Matrix x;
  x.init(2 * k + 3, 2 * k + 3);
  for (int i = 0; i < 2 * k + 3; i++) x.s[i][i] = 1;
  while (n) {
    if (n & 1) x = x * a;
    a = a * a;
    n /= 2;
  }
  return x;
}
int main() {
  init();
  cin >> n >> k;
  ans.init(2 * k + 3, 1);
  tmp.init(2 * k + 3, 2 * k + 3);
  for (int i = 0; i < 2 * k + 2; i++) ans.s[i][0] = 1;
  for (int i = 0; i <= k; i++) {
    for (int j = 0; j <= i; j++)
      tmp.s[i][j] = tmp.s[i + k + 1][j] = tmp.s[i][j + k + 1] = c[i][j];
  }
  tmp.s[2 * k + 2][k] = tmp.s[2 * k + 2][2 * k + 2] = 1;
  ans = qpow(tmp, n) * ans;
  cout << ans.s[2 * k + 2][0] << endl;
  return 0;
}