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;
struct pair_hash {
  template <class T1, class T2>
  std::size_t operator()(const std::pair<T1, T2> &p) const {
    auto h1 = std::hash<T1>{}(p.first);
    auto h2 = std::hash<T2>{}(p.second);
    return h1 ^ h2;
  }
};
map<long long, long long> mymap;
int main() {
  long long n, k;
  cin >> n >> k;
  if (n == k) {
    cout << "YES\n";
    for (long long i = 0; i < k; ++i) {
      cout << 1 << " ";
    }
    return 0;
  }
  vector<long long> vec;
  long long size = 0;
  for (long long i = 0; i < 32; ++i) {
    if ((1 << i) & n) {
      size++;
      mymap[(1 << i)]++;
    }
  }
  for (auto it : mymap)
    cerr << " - "
         << "it"
         << ": " << it.first << "," << it.second << endl;
  ;
  if (size > k) {
    cout << "NO";
    return 0;
  }
  for (auto it = mymap.begin(); it != mymap.end();) {
    if (size == k) break;
    if (it->first != 1) {
      mymap[it->first / 2] += 2;
      mymap[it->first]--;
      size--;
      size += 2;
      if (!mymap[it->first]) {
        auto search = mymap.find(it->first);
        mymap.erase(search);
      }
      it = mymap.begin();
      continue;
    }
    it++;
    if (size == k) break;
  }
  cerr << "--------------------" << endl;
  for (auto it : mymap)
    cerr << " - "
         << "it"
         << ": " << it.first << "," << it.second << endl;
  ;
  if (size != k) {
    cout << "NO";
    return 0;
  }
  cout << "YES\n";
  for (auto it : mymap)
    for (long long i = 0; i < it.second; ++i) {
      cout << it.first << " ";
    }
  return 0;
}