Problem: A positive integer x is called a power of two if it can be represented as x = 2^y, where y is a non-negative integer. So, the powers of two are 1, 2, 4, 8, 16, ....

You are given two positive integers n and k. Your task is to represent n as the sum of exactly k powers of two.

Input

The only line of the input contains two integers n and k (1 ≤ n ≤ 10^9, 1 ≤ k ≤ 2 ⋅ 10^5).

Output

If it is impossible to represent n as the sum of k powers of two, print NO.

Otherwise, print YES, and then print k positive integers b_1, b_2, ..., b_k such that each of b_i is a power of two, and ∑ _{i = 1}^{k} b_i = n. If there are multiple answers, you may print any of them.

Examples

Input


9 4


Output


YES
1 2 2 4 


Input


8 1


Output


YES
8 


Input


5 1


Output


NO


Input


3 7


Output


NO

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
int main() {
  int n, k;
  cin >> n >> k;
  map<int, int> ans;
  queue<int> q;
  for (int i = 0; i <= 30; i++)
    if (n & (1 << i)) {
      if (i > 0) q.push(1 << i);
      ans[1 << i]++;
    }
  if (int(ans.size()) > k) {
    puts("NO");
    return 0;
  }
  int cnt = ans.size();
  while (cnt < k && !q.empty()) {
    int z = q.front();
    q.pop();
    ans[z]--;
    ans[z / 2] += 2;
    if (z / 2 > 1) {
      q.push(z / 2);
      q.push(z / 2);
    }
    cnt++;
  }
  if (cnt < k) {
    puts("NO");
    return 0;
  }
  puts("YES");
  for (auto x : ans)
    for (int i = 0; i < x.second; i++) printf("%d ", x.first);
  puts("");
  return 0;
}