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 n, k;
  cin >> n >> k;
  if (k > n) {
    cout << "NO" << endl;
    return 0;
  }
  stack<long long> s;
  for (long long i = 0; i < 30; i++) {
    if ((n & (1 << i))) {
      s.push((1 << i));
    }
  }
  if (s.size() > k) {
    cout << "NO" << endl;
    return 0;
  }
  vector<long long> v;
  while (!s.empty()) {
    if (v.size() + s.size() == k) break;
    long long t = s.top();
    s.pop();
    if (t == 1)
      v.push_back(t);
    else {
      s.push(t / 2);
      s.push(t / 2);
    }
  }
  if (v.size() + s.size() < k) {
    cout << "NO" << endl;
    return 0;
  }
  cout << "YES" << endl;
  for (long long &x : v) {
    cout << x << " ";
  }
  while (!s.empty()) {
    cout << s.top() << " ";
    s.pop();
  }
  cout << endl;
  return 0;
}