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 r, c;
  long long a[90][90];
  Matrix(int r = 1, int c = 1) {
    this->r = r;
    this->c = c;
    memset(a, 0, sizeof a);
  }
  Matrix operator*(Matrix& b) {
    Matrix res(r, b.c);
    for (int k = 1; k <= c; k++) {
      for (int i = 1; i <= r; i++) {
        if (!a[i][k]) continue;
        for (int j = 1; j <= b.c; j++) {
          res.a[i][j] = (res.a[i][j] + a[i][k] * b.a[k][j]) % mod;
        }
      }
    }
    return res;
  }
  Matrix Power(long long t) {
    Matrix res(r, r), p = *this;
    for (int i = 1; i <= r; i++) res.a[i][i] = 1;
    while (t > 0) {
      if (t & 1) res = res * p;
      p = p * p;
      t >>= 1;
    }
    return res;
  }
  inline void Print() {
    printf("<%d,%d>\n", r, c);
    for (int i = 1; i <= r; i++) {
      for (int j = 1; j <= c; j++) {
        if (j == 1)
          printf("%d", a[i][j]);
        else
          printf(" %d", a[i][j]);
      }
      puts("");
    }
  }
};
long long C[45][45];
inline void init() {
  memset(C, 0, sizeof C);
  for (int i = 0; i <= 41; i++) C[i][0] = C[i][i] = 1;
  for (int i = 1; i <= 41; i++) {
    for (int j = 1; j < i; j++) {
      C[i][j] = (C[i - 1][j - 1] + C[i - 1][j]) % mod;
    }
  }
}
int main() {
  init();
  long long n;
  int k;
  scanf("%lld%d", &n, &k);
  int m = 2 * k + 3;
  Matrix f(m, m);
  f.a[1][1] = 1;
  for (int i = 2; i <= m; i++) {
    f.a[i][1] = 0;
    f.a[1][i] = C[k][(i - 2) % (k + 1)];
  }
  for (int i = 2; i <= k + 2; i++) {
    for (int j = 2; j <= i; j++) {
      f.a[i][j] = f.a[i][j + k + 1] = f.a[i + k + 1][j] = C[i - 2][j - 2];
    }
  }
  Matrix ans = f.Power(n - 1);
  long long sum = 0;
  for (int i = 1; i <= m; i++) sum = (sum + ans.a[1][i]) % mod;
  printf("%lld\n", sum);
  return 0;
}