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>
const int Maxk = 40;
const int Mod = 1000000007;
int C[Maxk + 5][Maxk + 5];
long long n;
int k;
void init() {
  C[0][0] = 1;
  for (int i = 1; i <= Maxk; i++) {
    C[i][0] = C[i][i] = 1;
    for (int j = 1; j < i; j++) {
      C[i][j] = (C[i - 1][j] + C[i - 1][j - 1]) % Mod;
    }
  }
}
struct Matrix {
  int a[2][2];
  void init() {
    a[0][0] = a[1][1] = 1;
    a[0][1] = a[1][0] = 0;
  }
  void clear() { a[0][0] = a[0][1] = a[1][0] = a[1][1] = 0; }
  friend Matrix operator*(Matrix a, Matrix b) {
    Matrix ans;
    for (int i = 0; i < 2; i++) {
      for (int j = 0; j < 2; j++) {
        ans.a[i][j] = 0;
        for (int k = 0; k < 2; k++) {
          ans.a[i][j] = (ans.a[i][j] + 1ll * a.a[i][k] * b.a[k][j]) % Mod;
        }
      }
    }
    return ans;
  }
  friend Matrix operator+(Matrix a, Matrix b) {
    Matrix ans;
    for (int i = 0; i < 2; i++) {
      for (int j = 0; j < 2; j++) {
        ans.a[i][j] = (a.a[i][j] + b.a[i][j]) % Mod;
      }
    }
    return ans;
  }
  friend Matrix operator*(Matrix a, int b) {
    Matrix ans;
    for (int i = 0; i < 2; i++) {
      for (int j = 0; j < 2; j++) {
        ans.a[i][j] = 1ll * a.a[i][j] * b % Mod;
      }
    }
    return ans;
  }
} fib, f[Maxk + 5], t[Maxk + 5], tmp;
int pow_m[Maxk + 5];
void solve(long long n) {
  if (n == 1) {
    for (int i = 0; i <= k; i++) {
      f[i] = fib;
    }
    tmp = fib;
    return;
  }
  long long m = (n >> 1);
  solve(m);
  for (int i = 0; i <= k; i++) {
    t[i] = f[i];
    f[i].clear();
  }
  pow_m[0] = 1;
  for (int i = 1, tmp_m = m % Mod; i <= k; i++) {
    pow_m[i] = 1ll * pow_m[i - 1] * tmp_m % Mod;
  }
  for (int i = 0; i <= k; i++) {
    for (int j = 0; j <= i; j++) {
      f[i] = f[i] + t[j] * (1ll * C[i][j] * pow_m[i - j] % Mod);
    }
    f[i] = f[i] * tmp;
  }
  for (int i = 0; i <= k; i++) {
    f[i] = f[i] + t[i];
  }
  tmp = tmp * tmp;
  if (n & 1) {
    tmp = tmp * fib;
    pow_m[0] = 1;
    for (int i = 1, tmp_m = n % Mod; i <= k; i++) {
      pow_m[i] = 1ll * pow_m[i - 1] * tmp_m % Mod;
    }
    for (int i = 0; i <= k; i++) {
      f[i] = f[i] + tmp * pow_m[i];
    }
  }
}
int main() {
  fib.a[0][0] = fib.a[0][1] = fib.a[1][0] = 1;
  fib.a[1][1] = 0;
  scanf("%lld%d", &n, &k);
  init();
  solve(n);
  printf("%d\n", f[k].a[0][0]);
  return 0;
}