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() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int n, k;
  cin >> n >> k;
  int bc = 0;
  vector<int> bts;
  for (int i = 0;; i++) {
    if ((1 << i) > n) break;
    if (n & (1 << i)) {
      bc++;
      bts.push_back(1 << i);
    }
  }
  if (bc > k || k > n) {
    cout << "NO" << endl;
    return 0;
  }
  int j = 0;
  k -= bc;
  int onec = 0;
  for (j = 0; j < bts.size(); j++) {
    if (k == 0) break;
    if (bts[j] == 1) {
      onec++;
      continue;
    }
    k--;
    bts.push_back(bts[j] / 2);
    bts.push_back(bts[j] / 2);
  }
  cout << "YES" << endl;
  for (int i = j; i < bts.size(); i++)
    if (bts[i] != 0) cout << bts[i] << " ";
  while (onec--) cout << "1 ";
  cout << endl;
  return 0;
}