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 int mod = 1e9 + 7;
long long int c[100][100];
struct Matrix {
  long long int m, n;
  vector<vector<long long int> > a;
  Matrix(long long int x, long long int y) {
    m = x;
    n = y;
    a.assign(x, vector<long long int>(y, 0));
  }
  Matrix operator*(Matrix b) {
    long long int i, k, j;
    Matrix ans(m, b.n);
    for (i = 0; i < m; i++)
      for (k = 0; k < n; k++)
        for (j = 0; j < b.n; j++) {
          ans.a[i][j] += a[i][k] * b.a[k][j];
          ans.a[i][j] %= mod;
        }
    return ans;
  }
  void prin() {
    long long int i, j;
    for (i = 0; i < m; i++) {
      for (j = 0; j < n; j++) {
        printf("%d ", a[i][j]);
      }
      printf("\n");
    }
  }
};
int main() {
  long long int k, i, j;
  long long int n;
  scanf("%I64d %I64d", &n, &k);
  Matrix a(1, k * 2 + 3), b(2 * k + 3, 2 * k + 3);
  for (i = 0; i <= k; i++) c[0][i] = c[i][i] = 1;
  for (i = 2; i <= k; i++)
    for (j = 1; j < i; j++) c[j][i] = (c[j][i - 1] + c[j - 1][i - 1]) % mod;
  for (i = 0; i <= k; i++) a.a[0][i] = 1;
  for (i = k + 1; i <= 2 * k + 1; i++) a.a[0][i] = 1;
  for (i = 0; i <= k; i++) {
    for (j = 0; j <= i; j++) {
      b.a[j][i + k + 1] = b.a[j + k + 1][i + k + 1] = b.a[j + k + 1][i] =
          c[j][i];
    }
  }
  b.a[2 * k + 2][2 * k + 2] = b.a[2 * k + 1][2 * k + 2] = 1;
  while (n > 0) {
    if (n % 2 == 1) a = a * b;
    b = b * b;
    n /= 2;
  }
  printf("%I64d", a.a[0][2 * k + 2]);
}