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;
void solve() {
  int n, k;
  cin >> n >> k;
  map<int, int> mp;
  for (int i = 0; i <= 32; i++) {
    if ((1LL << i) & n) mp[i] = 1;
  }
  int ans = __builtin_popcount(n);
  if ((ans > k) || (n < k))
    cout << "NO";
  else {
    while (ans != k) {
      for (auto a : mp) {
        if (a.first && a.second) {
          mp[a.first]--;
          mp[a.first - 1] += 2;
          ans++;
          break;
        }
      }
    }
    cout << "YES\n";
    for (auto a : mp) {
      for (int i = 0; i < a.second; i++) cout << (1LL << a.first) << " ";
    }
  }
  cout << "\n";
  ;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int t;
  t = 1;
  while (t--) {
    solve();
  }
  return 0;
}