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 long long mod = 1e9 + 7;
const int maxn = 100;
struct Matrix {
  long long m[maxn][maxn];
};
int N;
Matrix Mul(const Matrix &a, const Matrix &b) {
  Matrix c;
  memset(c.m, 0, sizeof(c.m));
  for (int i = 0; i < N; i++)
    for (int j = 0; j < N; j++)
      for (int k = 0; k < N; k++) {
        c.m[i][j] += (a.m[i][k] * b.m[k][j]) % mod;
        c.m[i][j] %= mod;
      }
  return c;
}
Matrix Mpow(Matrix a, long long n) {
  Matrix res;
  memset(res.m, 0, sizeof(res.m));
  for (int i = 0; i < N; i++) {
    res.m[i][i] = 1;
  }
  while (n) {
    if (n & 1) res = Mul(res, a);
    n >>= 1;
    a = Mul(a, a);
  }
  return res;
}
long long n;
int k;
long long C[50][50];
Matrix t;
void init() {
  C[0][0] = 1;
  C[1][0] = 1;
  C[1][1] = 1;
  for (int i = 2; i <= 40; i++) {
    for (int j = 0; j <= i; j++) {
      if (j == 0 || j == i)
        C[i][j] = 1;
      else
        C[i][j] = (C[i - 1][j - 1] + C[i - 1][j]) % mod;
    }
  }
  N = 2 * k + 3;
  t.m[0][0] = 1;
  for (int i = 1; i <= k + 1; i++) {
    t.m[0][i] = C[k][i - 1];
    t.m[0][i + k + 1] = C[k][i - 1];
  }
  for (int i = 1; i <= k + 1; i++) {
    for (int j = 1; j <= i; j++) {
      t.m[i][j] = C[i - 1][j - 1];
      t.m[i + k + 1][j] = C[i - 1][j - 1];
      t.m[i][j + k + 1] = C[i - 1][j - 1];
    }
  }
}
int main() {
  scanf("%lld%d", &n, &k);
  if (n == 1) {
    puts("1");
    return 0;
  }
  init();
  t = Mpow(t, n - 1);
  long long ans = 0;
  for (int i = 0; i < N; i++) {
    ans = (ans + t.m[0][i]) % mod;
  }
  printf("%lld\n", ans);
  return 0;
}