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 N = 101;
const int mod = 1e9 + 7;
struct matrix {
  long long mat[N][N];
  int r, c;
  matrix() {}
  matrix(int rr, int cc) { r = rr, c = cc; }
  void clear() { memset(mat, 0, sizeof mat); }
  void print() {
    puts("_______________");
    for (int i = 0; i < r; i++) {
      for (int j = 0; j < c; j++) {
        printf("%d ", mat[i][j]);
      }
      puts("");
    }
    puts("---------------");
  }
};
void MOD(long long &x) { x = (x % mod + mod) % mod; }
matrix operator*(const matrix &a, const matrix &b) {
  matrix c(a.r, b.c);
  c.clear();
  for (int k = 0; k < a.c; k++) {
    for (int i = 0; i < c.r; i++) {
      for (int j = 0; j < c.c; j++) {
        MOD(c.mat[i][j] += a.mat[i][k] * b.mat[k][j] % mod);
      }
    }
  }
  return c;
}
long long C[55][55], p[55];
void prework() {
  p[0] = 1;
  for (int i = 1; i <= 50; i++) p[i] = p[i - 1] * 2 % mod;
  for (int i = 0; i <= 50; i++) C[i][0] = 1;
  for (int i = 1; i <= 50; i++) {
    for (int j = 1; j <= i; j++)
      C[i][j] = (C[i - 1][j] + C[i - 1][j - 1]) % mod;
  }
}
long long solve(long long n, long long k) {
  if (n == 1) return 1;
  if (n == 2) return (p[k + 1] + 1) % mod;
  long long sum = 0;
  int kk = k * 2 + 3;
  matrix x(1, kk), y(kk, kk);
  x.mat[0][0] = (p[k + 1] + 1) % mod;
  for (int i = 0; i <= k; i++) {
    x.mat[0][i + 1] = p[i + 1];
    x.mat[0][i + k + 2] = 1;
  }
  y.mat[0][0] = 1;
  for (int i = 0; i <= k; i++) {
    y.mat[i + 1][0] = C[k][i];
    y.mat[i + k + 2][0] = C[k][i] * p[k - i] % mod;
    y.mat[i + 1][i + k + 2] = 1;
  }
  for (int i = 0; i <= k; i++) {
    for (int j = 0; j <= i; j++) {
      y.mat[j + 1][i + 1] = C[i][j];
      y.mat[j + k + 2][i + 1] = C[i][j] * p[i - j] % mod;
    }
  }
  n -= 2;
  for (; n; n >>= 1) {
    if (n & 1) x = x * y;
    y = y * y;
  }
  return x.mat[0][0];
}
int main() {
  long long n, k;
  prework();
  scanf("%lld%lld", &n, &k);
  printf("%lld\n", solve(n, k));
  return 0;
}