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 mod = 1e9 + 7;
int len;
long long n, k;
long long c[45][45];
struct matrix {
  long long m[100][100];
} mc, I;
matrix multi(matrix a, matrix b) {
  matrix ans;
  for (int i = 0; i < len; i++) {
    for (int j = 0; j < len; j++) {
      ans.m[i][j] = 0;
      for (int k = 0; k < len; k++) {
        ans.m[i][j] = (ans.m[i][j] + a.m[i][k] * b.m[k][j] % mod) % mod;
      }
    }
  }
  return ans;
}
matrix quick(matrix a, long long n) {
  matrix ans = I, t = a;
  while (n > 0) {
    if (n & 1) ans = multi(ans, t);
    t = multi(t, t);
    n >>= 1;
  }
  return ans;
}
int main() {
  for (int i = 0; i <= 40; i++) {
    c[i][0] = 1;
    for (int j = 1; j <= i; j++) {
      c[i][j] = c[i - 1][j - 1] + c[i - 1][j];
    }
  }
  for (int i = 0; i <= 40; i++) {
    I.m[i][i] = 1;
  }
  while (cin >> n >> k) {
    memset(mc.m, 0, sizeof(mc.m));
    len = 2 * k + 3;
    mc.m[0][0] = 1;
    for (int j = 0; j <= k; j++) {
      mc.m[0][j + 1] = c[k][j];
      mc.m[0][j + k + 2] = c[k][j];
    }
    mc.m[1][1] = mc.m[1][k + 2] = 1;
    for (int i = 1; i <= k; i++) {
      for (int j = 0; j <= i; j++) {
        mc.m[i + 1][j + 1] = c[i][j];
        mc.m[i + 1][j + k + 2] = c[i][j];
      }
    }
    mc.m[k + 2][1] = 1;
    for (int i = 1; i <= k; i++) {
      for (int j = 0; j <= i; j++) {
        mc.m[i + k + 2][j + 1] = c[i][j];
      }
    }
    mc = quick(mc, n - 1);
    long long ans = 0;
    for (int i = 0; i < len; i++) {
      ans = (ans + mc.m[0][i] % mod) % mod;
    }
    printf("%I64d\n", ans);
  }
  return 0;
}