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;
vector<pair<long long int, long long int> > r;
bool cmp(pair<long long int, long long int> a,
         pair<long long int, long long int> b) {
  return a.first < b.first || (a.first == b.first && a.second > b.second);
}
int main() {
  int n, k;
  cin >> n >> k;
  map<int, int> vys;
  queue<int> q;
  for (int i = 0; i <= 30; i++) {
    if (n & (1 << i)) {
      if (i > 0) {
        q.push(1 << i);
      }
      vys[1 << i]++;
    }
  }
  if (vys.size() > k) {
    cout << "NO" << endl;
    return 0;
  }
  int poc = vys.size();
  while (poc < k && !q.empty()) {
    int val = q.front();
    q.pop();
    vys[val]--;
    vys[val / 2] += 2;
    if (val / 2 > 1) {
      q.push(val / 2);
      q.push(val / 2);
    }
    poc++;
  }
  if (poc < k) {
    cout << "NO" << endl;
    return 0;
  }
  cout << "YES" << endl;
  for (auto it : vys) {
    for (int i = 0; i < it.second; i++) {
      cout << " " << it.first;
    }
  }
  cout << endl;
  return 0;
}