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 = 1000000007;
const int N = 100;
long long v[N][N], ans[N][N], t[N][N];
long long c[N][N];
void init(long long n, int k) {
  memset(c, 0, sizeof(c));
  memset(ans, 0, sizeof(ans));
  memset(v, 0, sizeof(v));
  for (int i = 0; i <= k; i++) c[i][0] = c[i][i] = 1;
  for (int i = 2; i <= k; i++) {
    for (int j = 1; j < i; j++) c[i][j] = (c[i - 1][j] + c[i - 1][j - 1]) % mod;
  }
  for (int i = 0; i <= k; i++) {
    for (int j = i; j <= k; j++) {
      v[i + k + 1][j] = v[i + k + 1][j + k + 1] = v[i][j + k + 1] = c[j][i];
    }
  }
  v[2 * k + 1][2 * k + 2] = 1;
  v[2 * k + 2][2 * k + 2] = 1;
  for (int i = 0; i < 2 * k + 3; i++) ans[i][i] = 1;
}
void mul(long long v1[][N], long long v2[][N], int n) {
  memset(t, 0, sizeof(t));
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < n; j++) {
      for (int k = 0; k < n; k++) {
        t[i][j] = (t[i][j] + v1[i][k] * v2[k][j] % mod) % mod;
      }
    }
  }
  memcpy(v1, t, sizeof(t));
}
void print(int k, long long v[][N]) {
  for (int i = 0; i < 2 * k + 3; i++) {
    for (int j = 0; j < 2 * k + 3; j++) cout << v[i][j] << "  ";
    cout << endl;
  }
}
long long solve(long long n, int k) {
  init(n, k);
  long long res = 0;
  while (n) {
    if (n & 1) mul(ans, v, 2 * k + 3);
    n = n >> 1;
    mul(v, v, 2 * k + 3);
  }
  for (int i = 0; i < 2 * k + 2; i++) res = (res + ans[i][2 * k + 2]) % mod;
  return res;
}
int main() {
  long long n;
  int k;
  while (cin >> n >> k) {
    long long ans = solve(n, k);
    cout << ans << endl;
  }
  return 0;
}