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;
long long power(long long x, long long y) {
  long long ans = 1;
  while (y) {
    if (y % 2) ans = ans * x;
    y /= 2;
    x = x * x;
  }
  return ans;
}
int main() {
  long long n, k;
  cin >> n >> k;
  vector<long long> v;
  vector<long long> ans;
  long long m = n, count = 0;
  while (m) {
    if (m & 1) ans.push_back(power(2, count));
    count++;
    m >>= 1;
  }
  if (ans.size() > k || k > n) {
    cout << "NO" << endl;
    return 0;
  }
  cout << "YES" << endl;
  while (ans.size() < k) {
    for (int it = 0; it < ans.size() && ans.size() < k; it++) {
      if (ans[it] != 1) {
        ans.push_back(ans[it] / 2);
        ans[it] /= 2;
      }
    }
  }
  for (int i = 0; i < k; i++) cout << ans[i] << " ";
}