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 N = 80 + 7;
const int MOD = 1e9 + 7;
struct mat {
  int val[N][N];
  int n;
  mat(int x, int v = 0) {
    n = x;
    for (int i = 0; i < n; i++)
      for (int j = 0; j < n; j++) val[i][j] = i == j ? v : 0;
  }
  int* operator[](int x) { return val[x]; }
};
mat operator*(mat a, mat b) {
  mat res(a.n, 0);
  for (int i = 0; i < a.n; i++)
    for (int j = 0; j < a.n; j++)
      for (int k = 0; k < a.n; k++)
        res[i][k] = (res[i][k] + 1LL * a[i][j] * b[j][k] % MOD) % MOD;
  return res;
}
mat operator^(mat a, long long b) {
  mat res(a.n, 1);
  while (b) {
    if (b & 1) res = res * a;
    a = a * a;
    b >>= 1;
  }
  return res;
}
int c[N][N];
int pw2[N];
int main() {
  ios::sync_with_stdio(false);
  long long n, k;
  cin >> n >> k;
  pw2[0] = 1;
  for (int i = 1; i < N; i++) pw2[i] = pw2[i - 1] * 2LL % MOD;
  for (int i = 0; i < N; i++) c[i][0] = 1;
  for (int i = 1; i < N; i++)
    for (int j = 1; j <= i; j++)
      c[i][j] = (c[i - 1][j] + c[i - 1][j - 1]) % MOD;
  mat a(2 * k + 3, 0);
  for (int i = 0; i <= k; i++) {
    for (int j = 0; j <= i; j++) {
      a[i][j] = c[i][j];
      a[i][j + k + 1] = 1LL * c[i][j] * pw2[i - j] % MOD;
    }
    a[i + k + 1][i] = 1;
  }
  a[2 * k + 2][2 * k + 2] = 1;
  a[2 * k + 2][k] = 1;
  a = a ^ (n + 1);
  cout << a[2 * k + 2][0] << "\n";
  return 0;
}