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;
int choose[42][42];
vector<vector<long long> > id(int n) {
  vector<vector<long long> > a(n, vector<long long>(n));
  for (auto i = (0); i < n; i++) a[i][i] = 1;
  return a;
}
vector<vector<long long> > operator*(vector<vector<long long> > a,
                                     vector<vector<long long> > b) {
  int n = a.size(), p = b.size(), m = b[0].size();
  vector<vector<long long> > c(n, vector<long long>(m));
  for (auto i = (0); i < n; i++)
    for (auto j = (0); j < m; j++)
      for (auto k = (0); k < p; k++)
        c[i][j] = ((c[i][j] + ((a[i][k] * b[k][j]) % 1000000007)) % 1000000007);
  return c;
}
vector<vector<long long> > operator^(vector<vector<long long> > a,
                                     long long b) {
  if (!b) return id(a.size());
  vector<vector<long long> > next = (a * a) ^ (b / 2);
  return b % 2 ? next * a : next;
}
int main() {
  for (auto n = (0); n < 42; n++) choose[n][0] = 1;
  for (auto n = (1); n < 42; n++)
    for (auto k = (1); k < n + 1; k++)
      choose[n][k] = ((choose[n - 1][k] + choose[n - 1][k - 1]) % 1000000007);
  long long t;
  int p;
  cin >> t >> p;
  int s = 2 * p + 3;
  vector<vector<long long> > m(s, vector<long long>(s)),
      x(s, vector<long long>(1));
  for (auto i = (0); i < p + 1; i++)
    for (auto j = (0); j < i + 1; j++)
      m[i][j] = m[i][j + p + 1] = m[i + p + 1][j] = choose[i][j];
  m[s - 1][p] = m[s - 1][s - 1] = 1;
  for (auto i = (0); i < p + 1; i++) x[i][0] = x[i + p + 1][0] = 1;
  cout << ((m ^ t) * x)[s - 1][0] << endl;
  return 0;
}