Problem: Numbers k-bonacci (k is integer, k > 1) are a generalization of Fibonacci numbers and are determined as follows:

  * F(k, n) = 0, for integer n, 1 ≤ n < k; 
  * F(k, k) = 1; 
  * F(k, n) = F(k, n - 1) + F(k, n - 2) + ... + F(k, n - k), for integer n, n > k. 



Note that we determine the k-bonacci numbers, F(k, n), only for integer values of n and k.

You've got a number s, represent it as a sum of several (at least two) distinct k-bonacci numbers. 

Input

The first line contains two integers s and k (1 ≤ s, k ≤ 109; k > 1).

Output

In the first line print an integer m (m ≥ 2) that shows how many numbers are in the found representation. In the second line print m distinct integers a1, a2, ..., am. Each printed integer should be a k-bonacci number. The sum of printed integers must equal s.

It is guaranteed that the answer exists. If there are several possible answers, print any of them.

Examples

Input

5 2


Output

3
0 2 3


Input

21 5


Output

3
4 1 16

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
int k, s;
const int MAXN = 1e6 + 10;
int F[MAXN];
int main() {
  memset(F, 0, sizeof(F));
  cin >> s >> k;
  int sum = 1, m = 0, t = 1;
  F[1] = 1;
  for (int i = 2;; i++) {
    F[i] = sum;
    if (sum > s) break;
    sum += F[i];
    if (i > k) m = F[i - k];
    sum -= m;
    t++;
  }
  vector<int> ans;
  while (s) {
    int cnt = upper_bound(F + 1, F + t + 1, s) - F;
    cnt--;
    s -= F[cnt];
    ans.push_back(F[cnt]);
  }
  printf("%d\n", ans.size() + 1);
  int flag = 1;
  printf("0 ");
  for (int i = 0; i < ans.size(); i++) {
    if (flag)
      flag = 0;
    else
      printf(" ");
    printf("%d", ans[i]);
  }
  puts("");
  return 0;
}