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(NULL);
  int n, k;
  cin >> n >> k;
  if (k <= n && k >= __builtin_popcount(n)) {
    cout << "YES\n";
    int c = __builtin_popcount(n);
    vector<int> ans;
    for (int i = 31; i >= 0; i--) {
      int z = n >> i;
      if (z & 1) ans.push_back(pow(2, i));
    }
    k -= c;
    int index = 0;
    while (k > 0) {
      if (ans[index] > 1) {
        ans.push_back(ans[index] / 2);
        ans.push_back(ans[index] / 2);
        ans[index] = -1;
        k--;
      }
      index++;
    }
    for (int i = 0; i < ans.size(); i++) {
      if (ans[i] == -1) continue;
      cout << ans[i] << " ";
    }
    cout << "\n";
  } else
    cout << "NO";
  return 0;
}