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 inf = 0x3f3f3f3f, N = 95, M = 4e1 + 5, MOD = 1e9 + 7;
int T, cas = 0;
long long t, n;
int k;
long long C[M][M];
struct Matrix {
  long long m[N][N];
  Matrix() { memset(m, 0, sizeof(m)); }
} A;
void init() {
  memset(C, 0, sizeof(C));
  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]) % MOD;
  }
}
Matrix multi(Matrix a, Matrix b) {
  Matrix c;
  for (int i = 0; i < (n); i++) {
    for (int j = 0; j < (n); j++) {
      for (int k = 0; k < (n); k++) {
        c.m[i][j] += a.m[i][k] * b.m[k][j];
        c.m[i][j] %= MOD;
      }
    }
  }
  return c;
}
Matrix power(Matrix base, long long t) {
  Matrix x;
  for (int i = 0; i < n; i++) x.m[i][i] = 1;
  while (t) {
    if (t & 1) x = multi(x, base);
    base = multi(base, base);
    t /= 2;
  }
  return x;
}
int main() {
  while (cin >> t >> k) {
    n = 2 * k + 3;
    init();
    Matrix ret;
    for (int i = (0); i <= (k); i++)
      for (int j = (0); j <= (i); j++)
        ret.m[i][j] = ret.m[i + k + 1][j] = ret.m[i][j + k + 1] = C[i][j];
    ret.m[2 * k + 2][k] = ret.m[2 * k + 2][2 * k + 2] = 1;
    Matrix Ans = power(ret, t);
    long long ans = 0;
    for (int i = 0; i < (2 * k + 2); i++) {
      ans += Ans.m[2 * k + 2][i];
      ans %= MOD;
    }
    cout << ans << endl;
  }
  return 0;
}