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 mo = 1000000000 + 7;
long long n;
int k, N;
long long C[50][50];
struct Matrix {
  long long a[200][200];
  int n, m;
  Matrix operator*(const Matrix &x) {
    Matrix T;
    memset(T.a, 0, sizeof(T.a));
    T.n = n, T.m = x.m;
    for (int i = 1; i <= T.n; i++)
      for (int j = 1; j <= T.m; j++)
        for (int k = 1; k <= m; k++)
          T.a[i][j] = (T.a[i][j] + a[i][k] * x.a[k][j]) % mo;
    return T;
  }
} G, Ans;
void prepare() {
  for (int i = 0; i <= k; 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]) % mo;
  }
}
Matrix pow(Matrix &a, long long b) {
  Matrix sum = a;
  b--;
  for (; b; b >>= 1) {
    if (b & 1) sum = sum * a;
    a = a * a;
  }
  return sum;
}
int main() {
  scanf("%I64d%d", &n, &k);
  N = 2 * k + 3;
  prepare();
  for (int i = 0; i <= k; i++)
    for (int j = 0; j <= i; j++)
      G.a[i + 1][j + 1] = G.a[i + 1 + k + 1][j + 1] =
          G.a[i + 1][j + k + 1 + 1] = C[i][j];
  G.a[N][k + 1] = G.a[N][N] = 1;
  G.n = G.m = N;
  for (int i = 1; i <= N - 1; i++) Ans.a[i][1] = 1;
  Ans.n = N, Ans.m = 1;
  Ans = pow(G, n) * Ans;
  printf("%I64d", Ans.a[N][1]);
  return 0;
}