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() {
  long long int n, k;
  cin >> n >> k;
  long long int count = 0;
  queue<long long int> q;
  long long int i = 0;
  map<long long int, long long int> mp;
  while (n) {
    if (n & 1) {
      count++;
      long long int b = pow(2, i);
      q.push(b);
      mp[b]++;
    }
    n >>= 1;
    i++;
  }
  if (count > k) {
    cout << "NO";
    return 0;
  }
  if (count == k) {
    cout << "YES" << endl;
    while (!q.empty()) {
      cout << q.front() << " ";
      q.pop();
    }
    return 0;
  }
  while (!q.empty() && count < k) {
    long long int a = q.front();
    q.pop();
    if (a == 1) {
      continue;
    }
    mp[a]--;
    a /= 2;
    mp[a] += 2;
    q.push(a);
    q.push(a);
    count += 1;
  }
  if (count < k) {
    cout << "NO";
    return 0;
  }
  map<long long int, long long int>::iterator it;
  cout << "YES" << endl;
  for (it = mp.begin(); it != mp.end(); it++) {
    for (int i = 0; i < it->second; i++) cout << it->first << " ";
  }
  return 0;
}