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 kbon[1000];
vector<int> v;
int main() {
  int s, k;
  scanf(" %d %d", &s, &k);
  kbon[0] = 1;
  kbon[1] = 1;
  int x = 2;
  while (kbon[x - 1] < s) {
    if (x - 1 - k >= 0)
      kbon[x] = kbon[x - 1] * 2 - kbon[x - 1 - k];
    else
      kbon[x] = kbon[x - 1] * 2;
    x++;
  }
  x--;
  while (s > 0) {
    if (kbon[x] <= s) {
      s -= kbon[x];
      v.push_back(kbon[x]);
    }
    x--;
  }
  printf("%d\n", v.size() + 1);
  printf("0");
  for (int i = v.size() - 1; i >= 0; i--) printf(" %d", v[i]);
  return 0;
}