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 n, m;
vector<int> bonac;
queue<int> now;
int i, j;
vector<int> hasil;
int sumnow;
int main() {
  scanf("%d %d", &n, &m);
  bonac.push_back(1);
  now.push(1);
  sumnow = 1;
  while (sumnow <= n) {
    now.push(sumnow);
    bonac.push_back(sumnow);
    sumnow += sumnow;
    while (now.size() > m) {
      sumnow -= now.front();
      now.pop();
    }
  }
  for (i = bonac.size() - 1; i >= 0; --i) {
    if (n >= bonac[i]) {
      n -= bonac[i];
      hasil.push_back(bonac[i]);
    }
  }
  if (hasil.size() == 1) hasil.push_back(0);
  printf("%d\n", hasil.size());
  for (i = 0; i < hasil.size(); ++i) printf("%d ", hasil[i]);
  printf("\n");
}