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, cnt = 0;
  cin >> n >> k;
  vector<int> v;
  for (int i = 0; i <= 31; i++) {
    if ((1 << i) & n) {
      v.push_back(1 << i);
    }
  }
  if (v.size() > k) {
    puts("NO\n");
    return 0;
  }
  while (v.size() < k) {
    bool ok = 0;
    for (int i = 0; i < v.size(); i++) {
      if (v[i] > 1) {
        v.push_back(v[i] / 2);
        v[i] /= 2;
        ok = 1;
        if (v.size() == k) {
          cout << "YES\n";
          for (int j = 0; j < v.size(); j++) {
            cout << v[j] << ' ';
          }
          return 0;
        }
      }
    }
    if (!ok) {
      puts("NO\n");
      return 0;
    }
  }
  if (k > v.size()) {
    puts("NO\n");
  } else {
    cout << "YES\n";
    for (int i = 0; i < v.size(); i++) {
      cout << v[i] << ' ';
    }
  }
  return 0;
}