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 inf = (1LL << 60) - 1;
int dx[] = {-1, 1, 0, 0};
int dy[] = {0, 0, -1, 1};
void solve() {
  long long n, k;
  cin >> n >> k;
  map<long long, long long> ans;
  queue<long long> q;
  for (long long i = 0; i <= 30; i++) {
    if (n & (1 << i)) {
      if (i > 0) q.push(1 << i);
      ans[1 << i]++;
    }
  }
  long long cnt = ans.size();
  if (cnt > k) {
    cout << "NO"
         << "\n";
    return;
  }
  while (cnt < k && !q.empty()) {
    long long z = q.front();
    q.pop();
    ans[z]--;
    ans[z / 2] += 2;
    if (z / 2 > 1) {
      q.push(z / 2);
      q.push(z / 2);
    }
    cnt++;
  }
  if (cnt < k) {
    cout << "NO"
         << "\n";
  } else {
    cout << "YES"
         << "\n";
    for (auto x : ans) {
      for (long long i = 0; i < x.second; i++) cout << x.first << " ";
    }
  }
}
int main() {
  ios::sync_with_stdio(0);
  cout.tie(0);
  ;
  solve();
  return 0;
}