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 n, k, t;
  cin >> n >> k;
  t = n;
  multiset<long long> a;
  multiset<long long>::iterator it, tempi;
  for (long long i = 30; i >= 0; i--) {
    if (n >= (1 << i)) {
      a.insert(1 << i);
      n %= (1 << i);
    }
  }
  if (k > t || k < a.size())
    cout << "NO";
  else {
    it = a.begin();
    cout << "YES \n";
    k -= a.size();
    while (k) {
      if (*it != 1) {
        k--;
        if (it != a.begin()) {
          it--;
          tempi = it;
          tempi++;
          a.insert(*(tempi) / 2);
          a.insert(*(tempi) / 2);
          a.erase(tempi);
        } else {
          tempi = it;
          a.insert(*(tempi) / 2);
          a.insert(*(tempi) / 2);
          a.erase(tempi);
          it = a.begin();
        }
      } else
        it++;
    }
    for (tempi = a.begin(); tempi != a.end(); tempi++) cout << *tempi << " ";
  }
}