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 matrix {
  long long x[100][100];
};
matrix multi(matrix a, matrix b) {
  matrix tmp;
  memset(tmp.x, 0, sizeof(tmp.x));
  for (int i = 1; i <= N; i++)
    for (int j = 1; j <= N; j++)
      for (int k = 1; k <= N; k++) {
        tmp.x[i][j] += a.x[i][k] * b.x[k][j];
        tmp.x[i][j] %= mod;
      }
  return tmp;
}
matrix ans;
void quick_multi(matrix a, long long k) {
  for (int i = 1; i <= N; i++)
    for (int j = 1; j <= N; j++) ans.x[i][j] = (i == j);
  while (k) {
    if (k & 1) ans = multi(ans, a);
    a = multi(a, a);
    k >>= 1;
  }
}
long long c[45][45];
void init(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;
}
int main() {
  long long n, k;
  while (~scanf("%lld%lld", &n, &k)) {
    if (n == 1) {
      printf("1\n");
      continue;
    }
    N = 2 * k + 3;
    init(k);
    matrix st;
    memset(st.x, 0, sizeof(st.x));
    st.x[1][1] = 1;
    for (int i = 0; i <= k; i++) st.x[1][i + 2] = c[k][i];
    for (int i = 0; i <= k; i++) st.x[1][k + i + 3] = c[k][i];
    for (int i = 0; i <= k; i++)
      for (int j = 0; j <= i; j++) {
        st.x[i + 2][j + 2] = c[i][j];
        st.x[i + 2][j + 3 + k] = c[i][j];
        st.x[i + 3 + k][j + 2] = c[i][j];
      }
    quick_multi(st, n - 1);
    long long res = 0;
    for (int i = 1; i <= N; i++) res = (res + ans.x[1][i]) % mod;
    printf("%lld\n", res);
  }
  return 0;
}