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 long long int inf = 1e9 + 7;
int cnt[32];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long int n, k;
  cin >> n >> k;
  queue<int> qq;
  int ctr = 0;
  for (int i = 0; i <= 30; i++) {
    if (n & (1 << i)) {
      if (i != 0) qq.push(i);
      cnt[i]++;
      ctr++;
    }
  }
  if (ctr > k) {
    cout << "NO\n";
    return 0;
  }
  while (ctr < k && !qq.empty()) {
    int tmp = qq.front();
    qq.pop();
    cnt[tmp--]--;
    cnt[tmp]++;
    cnt[tmp]++;
    if (tmp != 0) {
      qq.push(tmp);
      qq.push(tmp);
    }
    ctr++;
  }
  if (ctr < k) {
    cout << "NO\n";
    return 0;
  }
  cout << "YES\n";
  for (int i = 0; i <= 30; i++) {
    if (cnt[i] == 0) continue;
    for (int j = 0; j < cnt[i]; j++) cout << (1 << i) << " ";
  }
}