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;
long long fib[100002];
vector<long long> ans;
int main() {
  int s, k;
  scanf("%d %d", &s, &k);
  ans.push_back(0);
  if (k >= 30) {
    for (int i = 0; i <= 29; i++) {
      if ((1 << i) & s) ans.push_back((1 << i));
    }
  } else {
    fib[k] = 1;
    long long sum = 1, prev = 1, i;
    for (i = k + 1;; i++) {
      fib[i] = sum;
      if (fib[i] > 1000000000) break;
      sum -= fib[prev];
      prev++;
      sum += fib[i];
    }
    int sz = i;
    for (int i = sz; i >= k; i--) {
      if (fib[i] <= s) {
        s -= fib[i];
        ans.push_back(fib[i]);
      }
      if (s < (1 << k)) {
        break;
      }
    }
    if (s > 0) {
      for (int i = 0; i <= 29; i++) {
        if ((1 << i) & s) ans.push_back((1 << i));
      }
    }
  }
  printf("%d\n", ans.size());
  for (int i = 0; i < ans.size(); i++) {
    printf("%I64d ", ans[i]);
  }
}