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;
using ll = long long int;
int main() {
  int n, k;
  cin >> n >> k;
  multiset<int, greater<int>> ss;
  if (k > n) {
    cout << "NO" << endl;
  } else if (log2((double)n) - (int)log2(n) != 0 && k == 1) {
    cout << "NO" << endl;
  } else {
    bitset<32> bin(n);
    for (int i = 0; i < 32; i++) {
      if (bin[i]) {
        ss.emplace(1 << i);
      }
    }
    while (ss.size() < k && *ss.begin() > 1) {
      int num = *ss.begin();
      ss.erase(ss.begin());
      ss.emplace(num / 2);
      ss.emplace(num / 2);
    }
    if (ss.size() == k) {
      cout << "YES" << endl;
      for (auto x : ss) {
        cout << x << " ";
      }
    } else {
      cout << "NO" << endl;
    }
  }
}