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 mod = 1e9 + 7;
struct Matrix {
  int n, m;
  vector<vector<int> > a;
  Matrix(int x = 0, int y = 0, bool t = 0) {
    n = x;
    m = y;
    a.assign(n, vector<int>(m, 0));
    if (t)
      for (int i = 0; i < n; i++) a[i][i] = 1;
  }
  Matrix operator*(const Matrix &x) {
    const vector<vector<int> > &b = x.a;
    Matrix res(n, x.m);
    vector<vector<int> > &c = res.a;
    for (int i = 0; i < n; i++)
      for (int j = 0; j < x.m; j++)
        for (int k = 0; k < m; k++)
          c[i][j] = (c[i][j] + 1ll * a[i][k] * b[k][j]) % mod;
    return res;
  }
};
Matrix Pow(const Matrix &a, long long b) {
  if (b == 0) return Matrix(a.n, a.n, 1);
  if (b == 1) return a;
  Matrix t = Pow(a, b / 2);
  if (b % 2) return t * t * a;
  return t * t;
}
int c[100][100];
int main() {
  long long n;
  int k;
  cin >> n >> k;
  Matrix Unit(2 * k + 3, 2 * k + 3);
  for (int i = 0; i <= k; i++) c[i][0] = c[i][i] = 1;
  for (int i = 2; i <= k; i++)
    for (int j = 1; j < i; j++) c[i][j] = (c[i - 1][j] + c[i - 1][j - 1]) % mod;
  for (int i = 0; i <= k; i++)
    for (int j = 0; j <= i; j++)
      Unit.a[i][j + k + 1] = Unit.a[i + k + 1][j] =
          Unit.a[i + k + 1][j + k + 1] = c[i][j];
  Unit.a[2 * k + 2][2 * k + 1] = Unit.a[2 * k + 2][2 * k + 2] = 1;
  Matrix f0(2 * k + 3, 1);
  for (int i = 0; i <= 2 * k + 1; i++) f0.a[i][0] = 1;
  cout << (Pow(Unit, n) * f0).a[2 * k + 2][0];
}