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() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long n, k;
  cin >> n >> k;
  priority_queue<long long> ans;
  long long l = n;
  while (1) {
    long long i = 1 << (int)log2(n);
    if (n - i <= 0) {
      if (n - i == 0) {
        ans.push(i);
      }
      n -= i;
      break;
    } else {
      ans.push(i);
      n -= i;
    }
  }
  if (k < ans.size()) {
    cout << "NO";
  } else {
    if (k > l) {
      cout << "NO";
    } else {
      while (k != ans.size()) {
        long long ma = ans.top();
        if (ma > 1) {
          ans.pop();
          ans.push(ma / 2);
          ans.push(ma / 2);
        }
      }
      cout << "YES\n";
      while (!ans.empty()) {
        cout << ans.top() << " ";
        ans.pop();
      }
    }
  }
  return 0;
}