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;
struct node {
  long long a[100][100];
  node() { memset(a, 0, sizeof(a)); }
} p, v;
long long n, pre[60][60] = {1}, t, k, s = 0;
node operator*(node a, node b) {
  node t;
  for (int i = 0; i < n; i++)
    for (int j = 0; j < n; j++)
      for (int k = 0; k < n; k++)
        t.a[i][j] = (t.a[i][j] + a.a[i][k] * b.a[k][j]) % MOD;
  return t;
}
node pow(node a, long long b) {
  node t;
  for (int i = 0; i < n; i++) t.a[i][i] = 1;
  while (b > 0) {
    if (b & 1) t = t * a;
    b >>= 1;
    a = a * a;
  }
  return t;
}
int main() {
  cin >> t >> k;
  n = k + k + 3;
  for (int i = 0; i <= k; i++) {
    pre[i][0] = 1;
    for (int j = 1; j <= i; j++)
      pre[i][j] = (pre[i - 1][j - 1] + pre[i - 1][j]) % MOD;
  }
  for (int i = 0; i <= k; i++)
    for (int j = 0; j <= i; j++) {
      p.a[i][j] = pre[i][j];
      p.a[i + k + 1][j] = pre[i][j];
      p.a[i][j + k + 1] = pre[i][j];
    }
  p.a[n - 1][k] = p.a[n - 1][n - 1] = 1;
  v = pow(p, t);
  for (int i = 0; i < n - 1; i++) s = (s + v.a[n - 1][i]) % MOD;
  cout << s << endl;
  return 0;
}