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 P = 1000000007;
const int N = 100;
int k;
long long n;
struct mat {
  int x, y;
  long long v[N][N];
  mat() {
    x = y = 0;
    memset(v, 0, sizeof(v));
  }
  mat operator*(const mat &tmp) const {
    mat cur;
    cur.x = x, cur.y = tmp.y;
    for (int i = 1; i <= x; i++)
      for (int K = 1; K <= y; K++)
        for (int j = 1; j <= tmp.y; j++) {
          cur.v[i][j] = (cur.v[i][j] + v[i][K] * tmp.v[K][j] % P) % P;
        }
    return cur;
  }
} a, f;
long long C[N][N];
void prework() {
  C[0][0] = 1;
  for (int i = 1; i <= k; i++)
    for (int j = 0; j <= i; j++) {
      C[i][j] = C[i - 1][j];
      if (j > 0) C[i][j] = (C[i][j] + C[i - 1][j - 1]) % P;
    }
  a.x = 1, a.y = 2 * k + 3;
  long long cur = 1;
  for (int i = 0; i <= k; i++) {
    a.v[1][i + 1] = cur, a.v[1][(i + k + 1) + 1] = cur * 2 % P;
    cur = cur * 2 % P;
  }
  a.v[1][2 * k + 3] = 1;
  f.x = f.y = 2 * k + 3;
  for (int i = 0; i <= k; i++)
    for (int j = 0; j <= i; j++) f.v[(j + k + 1) + 1][i + 1] = C[i][j];
  for (int i = 0; i <= k; i++)
    for (int j = 0; j <= i; j++)
      f.v[j + 1][(i + k + 1) + 1] = f.v[(j + k + 1) + 1][(i + k + 1) + 1] =
          C[i][j];
  f.v[2 * k + 2][2 * k + 3] = f.v[2 * k + 3][2 * k + 3] = 1;
}
int main() {
  cin >> n >> k;
  prework();
  n--;
  while (n) {
    if (n & 1LL) a = a * f;
    f = f * f;
    n >>= 1;
  }
  printf("%lld\n", a.v[1][2 * k + 3]);
  return 0;
}