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 = 1e9 + 7;
int N;
struct node {
  long long a[100][100];
};
node shu, ans, mp;
node matrix(node x, node y) {
  for (int i = 1; i <= N; i++)
    for (int j = 1; j <= N; j++) {
      mp.a[i][j] = 0;
      for (int p = 1; p <= N; p++)
        mp.a[i][j] = (mp.a[i][j] + x.a[i][p] * y.a[p][j] + MOD) % MOD;
    }
  return mp;
}
void work(long long k) {
  for (int i = 1; i <= N; i++)
    for (int j = 1; j <= N; j++) ans.a[i][j] = 0;
  for (int i = 1; i <= N; i++) ans.a[i][i] = 1;
  node t = shu;
  while (k) {
    if (k & 1) ans = matrix(ans, t);
    k >>= 1;
    t = matrix(t, t);
  }
}
long long c[45][45];
void get_c(int k) {
  memset(c, 0, sizeof(c));
  c[0][0] = 1;
  for (int i = 1; i <= k; 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]) % MOD;
  }
}
void make(int k) {
  memset(shu.a, 0, sizeof(shu.a));
  shu.a[1][1] = 1;
  for (int i = 0; i <= k; i++) shu.a[1][i + 2] = c[k][i];
  for (int i = 0; i <= k; i++) shu.a[1][k + i + 3] = c[k][i];
  for (int i = 0; i <= k; i++)
    for (int j = 0; j <= i; j++) shu.a[i + 2][j + 2] = c[i][j];
  for (int i = 0; i <= k; i++)
    for (int j = 0; j <= i; j++) shu.a[i + 2][j + 3 + k] = c[i][j];
  for (int i = 0; i <= k; i++)
    for (int j = 0; j <= i; j++) shu.a[i + 3 + k][j + 2] = c[i][j];
}
long long n, k;
int main() {
  scanf("%I64d%I64d", &n, &k);
  if (n == 1) {
    printf("1\n");
  } else {
    N = 2 * k + 3;
    get_c(k);
    make(k);
    work(n - 1);
    long long ret = 0;
    for (int i = 1; i <= 2 * k + 3; i++) ret = (ret + ans.a[1][i]) % MOD;
    printf("%I64d\n", ret);
  }
  return 0;
}