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 maxn = 100005;
const int inf = (1 << 28) - 1;
const long long MOD = 1e9 + 7;
int Size;
struct Matrix {
  long long mat[105][105];
  void clear() { memset(mat, 0, sizeof(mat)); }
  void output() {
    for (int i = 0; i < Size; i++) {
      for (int j = 0; j < Size; j++) printf("%d ", mat[i][j]);
      printf("\n");
    }
  }
  Matrix operator*(const Matrix &b) const {
    Matrix ret;
    for (int i = 0; i < Size; i++)
      for (int j = 0; j < Size; j++) {
        ret.mat[i][j] = 0;
        for (int k = 0; k < Size; k++) {
          long long tmp = (long long)mat[i][k] * b.mat[k][j] % MOD;
          ret.mat[i][j] = (ret.mat[i][j] + tmp);
          if (ret.mat[i][j] >= MOD) ret.mat[i][j] -= MOD;
          if (ret.mat[i][j] < 0) ret.mat[i][j] += MOD;
        }
      }
    return ret;
  }
};
Matrix pow_M(Matrix a, long long n) {
  Matrix ret;
  ret.clear();
  for (int i = 0; i < Size; i++) ret.mat[i][i] = 1;
  Matrix tmp = a;
  while (n) {
    if (n & 1) ret = ret * tmp;
    tmp = tmp * tmp;
    n >>= 1;
  }
  return ret;
}
long long C[45][45];
void init() {
  C[0][0] = 1;
  for (int i = 1; i < 45; ++i) {
    C[i][0] = 1;
    C[i][1] = i;
    for (int j = 2; j <= i; ++j) C[i][j] = C[i - 1][j - 1] + C[i - 1][j];
  }
}
int main() {
  init();
  long long n, k;
  scanf("%lld%lld", &n, &k);
  if (n == 1) {
    printf("1\n");
    return 0;
  }
  Size = 2 * k + 3;
  Matrix A, B;
  A.clear();
  B.clear();
  A.mat[0][0] = 1;
  for (int i = 0; i <= k; ++i) A.mat[0][i + 2 + k] = A.mat[0][i + 1] = C[k][i];
  for (int i = 0; i <= k; ++i) {
    for (int j = 0; j <= i; ++j) {
      A.mat[i + 1][j + 1] = A.mat[i + 1][j + 2 + k] = C[i][j];
      A.mat[i + 2 + k][j + 1] = C[i][j];
    }
  }
  for (int i = 0; i < Size; ++i) B.mat[i][0] = 1;
  A = pow_M(A, n - 1);
  A = A * B;
  printf("%lld\n", A.mat[0][0]);
  return 0;
}