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 N = 105;
const int M = 45;
long long c[M][M];
struct Matrix {
  long long m[N][N];
};
Matrix A, I;
void init() {
  for (int i = 0; i < M; i++) {
    c[i][0] = c[i][i] = 1;
    if (i == 0) continue;
    for (int j = 1; j <= i; j++)
      c[i][j] = (c[i - 1][j] % mod + c[i - 1][j - 1] % mod) % mod;
  }
}
void initMatrix(int k) {
  int n = 2 * k + 3;
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < n; j++) {
      I.m[i][j] = (i == j);
    }
  }
  A.m[0][0] = 1;
  for (int i = 1; i < n; i++) {
    if (i <= k + 1)
      A.m[0][i] = c[k][k + 1 - i];
    else
      A.m[0][i] = A.m[0][i - k - 1];
  }
  for (int i = 1; i < n; i++) {
    A.m[i][0] = 0;
  }
  for (int i = 1; i < n; i++) {
    if (i <= k + 1) {
      for (int j = 1; j < n; j++) {
        if (j < i)
          A.m[i][j] = 0;
        else {
          if (j <= k + 1)
            A.m[i][j] = c[k + 1 - i][k + 1 - j];
          else
            A.m[i][j] = A.m[i][j - k - 1];
        }
      }
    } else {
      for (int j = 1; j < n; j++) {
        if (j < i - k - 1)
          A.m[i][j] = 0;
        else {
          if (j <= k + 1)
            A.m[i][j] = A.m[i - k - 1][j];
          else
            A.m[i][j] = 0;
        }
      }
    }
  }
}
Matrix multi(Matrix a, Matrix b, int n) {
  Matrix ret;
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < n; j++) {
      ret.m[i][j] = 0;
      for (int k = 0; k < n; k++) {
        ret.m[i][j] += a.m[i][k] * b.m[k][j];
        ret.m[i][j] %= mod;
      }
    }
  }
  return ret;
}
Matrix mul(Matrix A, int n, long long k) {
  Matrix ans = I, p = A;
  while (k) {
    if (k & 1) ans = multi(ans, p, n);
    k >>= 1;
    p = multi(p, p, n);
  }
  return ans;
}
int main() {
  init();
  long long n;
  int k;
  scanf("%lld%d", &n, &k);
  initMatrix(k);
  Matrix t = mul(A, 2 * k + 3, n - 1);
  long long ans = 0;
  for (int i = 0; i < 2 * k + 3; i++) {
    ans += t.m[0][i];
    ans %= mod;
  }
  cout << ans << endl;
  return 0;
}