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 P = 1000000007;
long long T, a[85][85], b[85][85], f[85], g[85];
int n;
void mul() {
  for (int i = 0; i < n; ++i)
    for (int j = 0; j < n; ++j) b[i][j] = 0;
  for (int k = 0; k < n; ++k)
    for (int i = 0; i < n; ++i)
      for (int j = 0; j < n; ++j) b[i][j] = (b[i][j] + a[i][k] * a[k][j]) % P;
  for (int i = 0; i < n; ++i)
    for (int j = 0; j < n; ++j) a[i][j] = b[i][j];
}
void mul1() {
  for (int i = 0; i < n; ++i) g[i] = 0;
  for (int i = 0; i < n; ++i)
    for (int j = 0; j < n; ++j) g[j] = (g[j] + f[i] * a[j][i]) % P;
  for (int i = 0; i < n; ++i) f[i] = g[i];
}
int main() {
  scanf("%I64d%d", &T, &n), ++n;
  for (int i = 0; i < n; ++i) {
    for (int j = 0; j < i + 1; ++j)
      a[i][j] = i && j ? (a[i - 1][j] + a[i - 1][j - 1]) % P : 1;
    for (int j = 0; j < n; ++j) a[i][j + n] = a[i + n][j] = a[i][j];
  }
  a[n * 2][n - 1] = a[n * 2][n * 2] = 1;
  n *= 2;
  for (int i = 0; i < n; ++i) f[i] = 1;
  f[n++] = 0;
  for (; T; T >>= 1, mul())
    if (T & 1) mul1();
  printf("%I64d\n", f[n - 1]);
  return 0;
}