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() {
  long long int n, k, d = 0;
  cin >> n >> k;
  long long int m = 1;
  vector<long long int> v(0);
  while (m <= n) {
    if ((m & n) == m) {
      ++d;
      v.push_back(m);
    }
    m = m << 1;
  }
  if (d > k) {
    cout << "NO";
  } else {
    long long int h = 0, t = 0;
    while (v.size() < k && t == 0) {
      if (h != v.size() && v[h] != 1 && v[h] != 0) {
        v[h] = v[h] / 2;
        v.push_back(v[h]);
      } else {
        if (h == v.size()) {
          cout << "NO" << endl;
          ++t;
        } else {
          ++h;
        }
      }
    }
    if (t == 0) {
      cout << "YES" << endl;
      for (long long int u = 0; u < v.size(); ++u) {
        cout << v[u] << " ";
      }
    }
  }
}