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 = 1000000007;
struct ty {
  long long a[90][90];
} rt, m;
int k, p;
;
long long n, c[45][45];
void pre(int k) {
  c[0][0] = 1;
  for (int i = 1; i <= k; i++) {
    c[0][i] = 1;
    for (int j = 1; j <= k; j++)
      c[j][i] = (c[j - 1][i - 1] + c[j][i - 1]) % mod;
  }
}
ty multi(ty x, ty y) {
  ty mn;
  for (int i = 0; i <= p; i++)
    for (int j = 0; j <= p; j++) {
      long long d = 0;
      for (int k = 0; k <= p; k++)
        d = (d + (long long)x.a[i][k] * y.a[k][j] % mod) % mod;
      mn.a[i][j] = d;
    }
  return mn;
}
ty quick(long long y) {
  ty s = rt;
  ty t = m;
  while (y != 0) {
    if (y & 1) {
      s = multi(t, s);
    }
    y >>= 1;
    t = multi(t, t);
  }
  return s;
}
int main() {
  scanf("%I64d%d", &n, &k);
  pre(k);
  for (int i = 0; i <= k; i++) {
    for (int j = 0; j <= i; j++) m.a[i][j] = c[j][i];
    for (int j = k + 1; j <= k + 1 + i; j++) m.a[i][j] = m.a[i][j - k - 1];
  }
  for (int i = k + 1; i <= k + k + 1; i++)
    for (int j = 0; j <= k; j++) m.a[i][j] = m.a[i - k - 1][j];
  m.a[k + k + 2][k] = 1;
  m.a[k + k + 2][k + k + 2] = 1;
  p = k + k + 2;
  for (int i = 0; i < p; i++) rt.a[i][0] = 1;
  ty ans = quick(n);
  printf("%I64d", ans.a[p][0]);
  return 0;
}