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;
const int N = 2e5 + 5, OO = 0x3f3f3f3f;
int n, k;
void _run() {
  queue<int> q;
  map<int, int> freq;
  scanf("%d %d", &n, &k);
  for (int i = 0; i <= 30; ++i) {
    if (n & (1 << i)) {
      if (i > 0) q.push(1 << i);
      freq[1 << i]++;
    }
  }
  if (freq.size() == k) {
    puts("YES");
    for (auto x : freq) {
      for (int i = 0; i < x.second; ++i) printf("%d ", x.first);
    }
    puts("");
    return;
  }
  if (freq.size() > k) {
    puts("NO");
    return;
  }
  int cnt = freq.size();
  while (cnt < k && !q.empty()) {
    int u = q.front();
    q.pop();
    freq[u]--;
    freq[u / 2] += 2;
    if (u / 2 > 1) {
      q.push(u / 2);
      q.push(u / 2);
    }
    cnt++;
  }
  if (cnt < k)
    puts("NO");
  else {
    puts("YES");
    for (auto x : freq) {
      for (int i = 0; i < x.second; ++i) printf("%d ", x.first);
    }
    puts("");
  }
}
int main() {
  int tc = 1;
  while (tc--) _run();
  return 0;
}