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;
struct mat {
  long long a[90][90];
  int r, c;
  mat operator*(const mat &b) const {
    mat ret;
    for (int i = 0; i < r; i++) {
      for (int j = 0; j < b.c; j++) {
        ret.a[i][j] = 0;
        for (int k = 0; k < c; k++)
          ret.a[i][j] += a[i][k] * b.a[k][j], ret.a[i][j] %= MOD;
      }
    }
    ret.r = r;
    ret.c = b.c;
    return ret;
  }
  void init_unit(int x) {
    r = c = x;
    for (int i = 0; i < r; i++) {
      for (int j = 0; j < c; j++) {
        if (i == j)
          a[i][j] = 1;
        else
          a[i][j] = 0;
      }
    }
  }
} unit;
mat qmod(mat p, long long n) {
  unit.init_unit(p.c);
  mat ans = unit;
  while (n) {
    if (n & 1) ans = p * ans;
    p = p * p;
    n >>= 1;
  }
  return ans;
}
int k;
long long n, c[45][45];
void init() {
  memset(c, 0, sizeof(c));
  c[0][0] = 1;
  for (int i = 1; i <= 40; i++) {
    c[i][0] = c[i][i] = 1;
    for (int j = 1; j < i; j++) {
      c[i][j] = (c[i - 1][j - 1] + c[i - 1][j]) % MOD;
    }
  }
}
mat A, B;
int main() {
  init();
  while (~scanf("%lld%d", &n, &k)) {
    if (n == 1)
      printf("1\n");
    else {
      A.r = A.c = 2 * k + 3;
      memset(A.a, 0, sizeof(A.a));
      memset(B.a, 0, sizeof(B.a));
      A.a[0][0] = A.a[0][k + 1] = 1;
      for (int i = 0; i <= k; i++) {
        for (int j = 0; j <= i; j++) {
          A.a[i + 1][j + 1] = c[i][j];
          A.a[i + 1][j + k + 2] = (1LL << (i - j)) % MOD * c[i][j] % MOD;
        }
      }
      for (int i = k + 2; i <= 2 * k + 2; i++) A.a[i][i - (k + 1)] = 1;
      B.r = 2 * k + 3;
      B.c = 1;
      B.a[0][0] = 1;
      for (int i = 1; i <= k + 1; i++) B.a[i][0] = (1LL << i) % MOD;
      for (int i = k + 2; i <= 2 * k + 2; i++) B.a[i][0] = 1;
      A = qmod(A, n - 2);
      A = A * B;
      printf("%lld\n", (A.a[0][0] + A.a[k + 1][0]) % MOD);
    }
  }
  return 0;
}