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;
vector<int> v;
priority_queue<int> q;
void init() {
  v.push_back(1);
  int t = 1;
  while (1) {
    t *= 2;
    if (t <= 1e9)
      v.push_back(t);
    else
      break;
  }
  sort(v.rbegin(), v.rend());
}
int main() {
  int n, k;
  cin >> n >> k;
  init();
  int cnt = 0;
  for (int i = 0; i < v.size(); i++) {
    while (v[i] <= n) {
      n -= v[i];
      q.push(v[i]);
      cnt++;
    }
    if (!n) break;
  }
  if (cnt > k) return puts("NO"), 0;
  int t = k - cnt;
  while (!q.empty()) {
    if (t == 0) break;
    int x = q.top();
    q.pop();
    if (x == 1) return puts("NO"), 0;
    x /= 2;
    q.push(x);
    q.push(x);
    t--;
  }
  puts("YES");
  while (!q.empty()) {
    cout << q.top() << " ";
    q.pop();
  }
  cout << endl;
  return 0;
}