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;
long long a[100][100], b[100][1], C[50][50];
long long n, k;
void cal() {
  for (int i = 0; i <= k; ++i) {
    C[i][0] = 1;
    for (int j = 1; j <= i; ++j)
      C[i][j] = (C[i - 1][j - 1] + C[i - 1][j]) % MOD;
  }
}
void init() {
  cal();
  for (int i = 0; i < 2 * k + 3; ++i) b[i][0] = 1;
  a[2 * k + 2][2 * k + 2] = 1;
  for (int i = 0; i < k + 1; ++i)
    a[2 * k + 2][i] = C[k][i], a[2 * k + 2][i + k + 1] = C[k][i];
  for (int i = 0; i < k + 1; ++i) {
    for (int j = i; j < k + 1; ++j)
      a[i][j] = a[i][j + k + 1] = a[i + k + 1][j] = C[k - i][j - i];
  }
}
void calmatrix1() {
  long long c[100][1] = {0};
  for (int i = 0; i < 2 * k + 3; ++i)
    for (int j = 0; j < 2 * k + 3; ++j)
      c[i][0] = (c[i][0] + a[i][j] * b[j][0]) % MOD;
  memcpy(b, c, sizeof(c));
}
void calmatrix2() {
  long long c[100][100] = {0};
  for (int i = 0; i < 2 * k + 3; ++i)
    for (int j = 0; j < 2 * k + 3; ++j)
      for (int t = 0; t < 2 * k + 3; ++t)
        c[i][t] = (c[i][t] + a[i][j] * a[j][t]) % MOD;
  memcpy(a, c, sizeof(c));
}
long long solve(long long x) {
  init();
  x -= 1;
  while (x) {
    if (x & 1) calmatrix1();
    calmatrix2();
    x >>= 1;
  }
  return b[2 * k + 2][0];
}
int main() {
  cin >> n >> k;
  cout << solve(n);
  return 0;
}