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;
const long long int maxn = 1e15;
const long long int mod = 1e9 + 7;
using namespace std;
int main() {
  long long int n, k;
  cin >> n >> k;
  long long int c = 0;
  vector<long long int> v;
  queue<long long int> q;
  map<long long int, long long int> m;
  long long int i;
  map<long long int, long long int>::iterator it;
  for (int i = 0; i <= 30; i++) {
    if (n & (1 << i)) {
      if (i > 0) q.push(1 << i);
      m[1 << i]++;
      c++;
    }
  }
  if (c > k)
    cout << "NO\n";
  else if (c == k) {
    cout << "YES\n";
    for (it = m.begin(); it != m.end(); it++) {
      while (it->second > 0) {
        cout << it->first << " ";
        it->second--;
      }
    }
  } else {
    while (c < k && !q.empty()) {
      int no = q.front();
      q.pop();
      m[no]--;
      m[no / 2] = m[no / 2] + 2;
      if ((no / 2) > 1) {
        q.push(no / 2);
        q.push(no / 2);
      }
      c++;
    }
    if (c < k)
      cout << "NO\n";
    else {
      cout << "YES\n";
      for (it = m.begin(); it != m.end(); it++) {
        while (it->second > 0) {
          cout << it->first << " ";
          it->second--;
        }
      }
    }
  }
  cerr << "\nTime elapsed:" << 1000 * clock() / CLOCKS_PER_SEC << "ms\n";
  return 0;
}