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;
inline void solve() {
  priority_queue<int> q;
  int n, k, i = 0;
  cin >> n >> k;
  while (n) {
    if ((n >> i) & 1) {
      int p = pow(2, i);
      q.push(p);
    }
    i++;
    if (i == 32) break;
  }
  while (q.size() < k) {
    int t = q.top();
    if (t & 1) {
      cout << "NO" << endl;
      return;
    }
    q.pop();
    q.push(t / 2);
    q.push(t / 2);
  }
  if (q.size() == k) {
    cout << "YES" << endl;
    while (q.size()) {
      cout << q.top() << " ";
      q.pop();
    }
  } else
    cout << "NO" << endl;
  return;
}
int main() {
  ios_base::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  int n = 1;
  while (n--) solve();
  ;
  return 0;
}