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;
  if (n < k) cout << "NO", exit(0);
  int cur = 0;
  vector<int> res;
  while (n) {
    int cur = 1;
    while (cur <= n) cur <<= 1;
    cur >>= 1;
    res.push_back(cur);
    n -= cur;
  }
  vector<int> f;
  int u = k, uy = -1;
  while (k) {
    if (k == res.size()) {
      f = res;
      break;
    }
    int W = 0;
    for (int i = 0; i < res.size(); i++) {
      int z = 1;
      int w = 0;
      while (z <= (k - ((int)res.size() - 1 - i)) && res[i]) {
        w++;
        z *= 2;
        res[i] /= 2;
      }
      if (w) {
        res[i] *= 2;
        if (!res[i]) res[i] = 1;
        z /= 2;
      }
      W |= (w > 1);
      k -= z;
      for (int j = 0; j < z; j++) f.push_back(res[i]);
    }
    if (!k) break;
    if (!W || f.size() > u) cout << "NO", exit(0);
    uy = k;
    k = u;
    res = f;
    f.clear();
  }
  cout << "YES" << endl;
  for (int i = 0; i < f.size(); i++) cout << f[i] << " ";
}