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() {
  int n, k;
  cin >> n >> k;
  int f = 1, d = 0;
  while (f <= n) {
    f *= 2;
    d++;
  }
  d -= 1;
  vector<int> vi;
  for (int i = 0; i <= d; i++) {
    if (n & (1 << i)) {
      vi.push_back(i);
    }
  }
  sort(vi.begin(), vi.end());
  if (vi.size() > k)
    cout << "NO";
  else {
    int l = vi.size();
    int f = l;
    while (l < k) {
      for (int i = vi.size() - 1; i >= 0; i--) {
        if (vi[i] > 0 && l < k) {
          vi[i]--;
          vi.push_back(vi[i]);
          l++;
        }
      }
      if (f == l) {
        cout << "NO";
        return 0;
      }
      f = l;
      sort(vi.begin(), vi.end());
    }
    if (l == k) {
      cout << "YES" << endl;
      for (int i = 0; i < vi.size(); i++) cout << (1 << vi[i]) << " ";
    } else
      cout << "NO";
  }
  return 0;
}