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 maxn = 85;
const int mod = 1000000007;
struct Matrix {
  long long mat[maxn][maxn];
  int n;
  Matrix(){};
  Matrix(int nn) { n = nn; }
  void Init() {
    for (int i = 0; i < n; ++i)
      for (int j = 0; j < n; ++j) mat[i][j] = (i == j);
  }
  void clear() { memset(mat, 0, sizeof(mat)); }
};
Matrix operator*(const Matrix& a, const Matrix& b) {
  Matrix c(a.n);
  c.clear();
  for (int k = 0; k < c.n; ++k)
    for (int i = 0; i < c.n; ++i)
      for (int j = 0; j < c.n; ++j)
        c.mat[i][j] = (c.mat[i][j] + a.mat[i][k] * b.mat[k][j]) % mod;
  return c;
}
long long C[55][55], p[55], num[maxn];
void Init() {
  memset(C, 0, sizeof(C));
  p[0] = 1;
  for (int i = 1; i <= 50; ++i) p[i] = (p[i - 1] * 2) % mod;
  for (int i = 0; i <= 50; ++i) C[i][0] = 1, C[i][1] = i;
  for (int i = 1; i <= 50; ++i) {
    for (int j = 2; j <= i; ++j)
      C[i][j] = (C[i - 1][j - 1] + C[i - 1][j]) % mod;
  }
}
long long solve(long long n, int k) {
  if (n == 1) return 1;
  if (n == 2) return (p[k + 1] + 1) % mod;
  long long sum = 0;
  int nn = 2 * k + 3;
  num[0] = p[k + 1] + 1;
  num[1] = 2;
  num[k + 2] = 1;
  for (int i = 1; i <= k; ++i) {
    num[i + 1] = 2 * p[i];
    num[i + k + 2] = 1;
  }
  Matrix x(nn), y(nn);
  x.Init();
  y.clear();
  y.mat[0][0] = 1;
  y.mat[1][1] = y.mat[k + 2][1] = 1;
  for (int i = 0; i <= k; ++i) {
    y.mat[i + 1][0] = C[k][i];
    y.mat[i + k + 2][0] = (C[k][i] * p[k - i]) % mod;
    y.mat[i + 1][i + k + 2] = 1;
  }
  for (int i = 0; i <= k; ++i) {
    for (int j = 0; j <= i; ++j) {
      y.mat[j + 1][i + 1] = C[i][j];
      y.mat[j + k + 2][i + 1] = (C[i][j] * p[i - j]) % mod;
    }
  }
  n -= 2;
  while (n) {
    if (n & 1) x = x * y;
    y = y * y;
    n >>= 1;
  }
  for (int i = 0; i < nn; ++i) sum = (sum + num[i] * x.mat[i][0]) % mod;
  return sum;
}
int main() {
  long long n;
  int k;
  Init();
  while (cin >> n >> k) {
    cout << solve(n, k) << endl;
  }
  return 0;
}