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 N = 1e5 + 50;
long long n, k;
vector<long long> ans;
vector<int> bit;
void Get(int x) {
  while (x) bit.push_back(x % 2), x /= 2;
}
int main() {
  cin >> n >> k;
  Get(n);
  if (k > n || k < count(bit.begin(), bit.end(), 1)) return puts("NO");
  for (int i = 0; i < bit.size(); ++i) {
    if (bit[i]) ans.push_back(1ll << i);
  }
  for (int i = 0; i < ans.size(); ++i) {
    if (ans.size() == k) break;
    while (ans[i] > 1) {
      if (ans.size() == k) break;
      ans.push_back(ans[i] / 2);
      ans[i] /= 2;
    }
  }
  sort(ans.begin(), ans.end());
  puts("YES");
  for (auto i : ans) cout << i << " ";
  return 0;
}