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 = 200;
const long long MOD = (int)1e9 + 7;
long long C[MAXN][MAXN], MatrixSize;
struct Matrix {
  long long a[MAXN][MAXN];
  void Init() {
    memset(a, 0, sizeof(a));
    for (int i = 0; i < MAXN; i++) a[i][i] = 1;
  }
  Matrix operator*(const Matrix &b) const {
    Matrix temp;
    for (int i = 0; i < MatrixSize; i++) {
      for (int j = 0; j < MatrixSize; j++) {
        temp.a[i][j] = 0;
        for (int k = 0; k < MatrixSize; k++) {
          temp.a[i][j] = (temp.a[i][j] + (a[i][k] * b.a[k][j]) % MOD) % MOD;
        }
      }
    }
    return temp;
  }
};
Matrix qpow(Matrix a, long long b) {
  Matrix ans;
  ans.Init();
  while (b) {
    if (b & 1) ans = ans * a;
    a = a * a;
    b >>= 1;
  }
  return ans;
}
void Init() {
  memset(C, 0, sizeof(C));
  for (int i = 0; i < MAXN; i++) {
    C[i][0] = 1;
    for (int j = 1; j <= i; j++)
      C[i][j] = (C[i - 1][j] + C[i - 1][j - 1]) % MOD;
  }
}
int main() {
  Init();
  long long n, k;
  scanf("%lld %lld", &n, &k);
  MatrixSize = 2 * k + 3;
  Matrix a;
  memset(a.a, 0, sizeof(a.a));
  a.a[0][0] = 1;
  for (int i = 0; i <= k; i++) a.a[0][i + 1] = a.a[0][i + k + 2] = C[k][i];
  for (int i = 0; i <= k; i++) {
    for (int j = 0; j <= i; j++) {
      a.a[i + 1][j + k + 2] = a.a[i + 1][j + 1] = a.a[i + k + 2][j + 1] =
          C[i][j];
    }
  }
  a = qpow(a, n - 1);
  Matrix s;
  for (int i = 0; i < MatrixSize; i++) s.a[i][0] = 1;
  s = a * s;
  printf("%lld\n", s.a[0][0]);
  return 0;
}