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 mod = 1e9 + 7;
int len;
long long n, k;
vector<vector<long long> > c(45, vector<long long>(45));
vector<vector<long long> > mc(85, vector<long long>(85));
vector<vector<long long> > I(85, vector<long long>(85));
vector<vector<long long> > operator*(vector<vector<long long> > a,
                                     vector<vector<long long> > b) {
  vector<vector<long long> > ans(85, vector<long long>(85));
  for (int i = 0; i < len; ++i) {
    for (int j = 0; j < len; ++j) {
      for (int k = 0; k < len; ++k) {
        ans[i][j] = (ans[i][j] + a[i][k] * b[k][j] % mod) % mod;
      }
    }
  }
  return ans;
}
vector<vector<long long> > ksm(vector<vector<long long> > a, long long n) {
  vector<vector<long long> > ans = I, t = a;
  while (n) {
    if (n & 1) ans = ans * t;
    t = t * t;
    n >>= 1;
  }
  return ans;
}
int main() {
  c[0][0] = 1;
  c[1][0] = c[1][1] = 1;
  for (int i = 2; i <= 40; i++) {
    for (int j = 0; j <= i; j++) {
      if (j == 0 || j == i)
        c[i][j] = 1;
      else
        c[i][j] = c[i - 1][j - 1] + c[i - 1][j];
    }
  }
  for (int i = 0; i <= 40; i++) {
    I[i][i] = 1;
  }
  while (cin >> n >> k) {
    mc.assign(85, vector<long long>(85, 0));
    len = 2 * k + 3;
    mc[0][0] = 1;
    for (int j = 0; j <= k; j++) {
      mc[0][j + 1] = c[k][j];
      mc[0][j + k + 2] = c[k][j];
    }
    mc[1][1] = mc[1][k + 2] = 1;
    for (int i = 1; i <= k; i++) {
      for (int j = 0; j <= i; j++) {
        mc[i + 1][j + 1] = c[i][j];
        mc[i + 1][j + k + 2] = c[i][j];
      }
    }
    mc[k + 2][1] = 1;
    for (int i = 1; i <= k; i++) {
      for (int j = 0; j <= i; j++) {
        mc[i + k + 2][j + 1] = c[i][j];
      }
    }
    mc = ksm(mc, n - 1);
    long long ans = 0;
    for (int i = 0; i < len; i++) {
      ans = (ans + mc[0][i] % mod) % mod;
    }
    printf("%lld\n", ans);
  }
  return 0;
}