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;
void g(int t) {
  if (t) {
    g((t - 1) / 26);
    putchar(65 + (t - 1) % 26);
  }
}
const int Mod = 1e9 + 7;
const int M = 45;
vector<vector<long long> > mul(vector<vector<long long> > &A,
                               vector<vector<long long> > &B) {
  vector<vector<long long> > C(A.size(), vector<long long>(B[0].size()));
  for (int i = 0; i < A.size(); i++) {
    for (int k = 0; k < B.size(); k++) {
      for (int j = 0; j < B[0].size(); j++)
        C[i][j] = (C[i][j] + A[i][k] * B[k][j]) % Mod;
    }
  }
  return C;
}
vector<vector<long long> > pow(vector<vector<long long> > A, long long int n) {
  vector<vector<long long> > B(A.size(), vector<long long>(A.size()));
  for (int i = 0; i < A.size(); i++) B[i][i] = 1;
  while (n > 0) {
    if (n & 1) B = mul(B, A);
    A = mul(A, A);
    n >>= 1;
  }
  return B;
}
int vis[300][300];
int n, k;
long long int s, c[M][M];
void init() {
  scanf("%I64d", &s);
  scanf("%d", &k);
  n = k * 2 + 3;
  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];
  }
}
int main() {
  init();
  vector<vector<long long> > base(100, vector<long long>(100));
  for (int i = 0; i <= k; i++) {
    for (int j = 0; j <= i; j++) {
      base[i][j] = base[i + k + 1][j] = base[i][j + k + 1] = c[i][j];
    }
  }
  base[2 * k + 2][k] = base[2 * k + 2][2 * k + 2] = 1;
  vector<vector<long long> > ans(100, vector<long long>(100));
  ans = pow(base, s);
  long long int g = 0;
  for (int i = 0; i < 2 * k + 2; i++) {
    g = (g + ans[2 * k + 2][i]) % Mod;
  }
  printf("%I64d\n", g);
}