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;
const int maxn = (int)1e5 + 10;
int n, k, ind, x;
vector<int> v;
int main() {
  ios_base::sync_with_stdio(false), cin.tie(NULL), cout.tie(NULL);
  cin >> n >> k;
  x = n;
  for (int i = 30; i >= 0; i--) {
    if (n >= (1 << i)) {
      n -= (1 << i);
      v.push_back(i);
    }
  }
  if (k > x) {
    cout << "NO";
    return cout << '\n', 0;
  }
  if (v.size() > k) {
    cout << "NO";
    return cout << '\n', 0;
  }
  while (v.size() != k) {
    if (v[ind] > 0) {
      v[ind]--;
      v.push_back(v[ind]);
    }
    ind++;
    if (ind >= v.size()) ind = 0;
  }
  cout << "YES" << '\n';
  for (int i = 0; i < v.size(); i++) cout << (1 << v[i]) << " ";
  return cout << '\n', 0;
}