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;
int32_t main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  long long n, k;
  cin >> n >> k;
  vector<long long> v;
  long long x, f = 0;
  while (n && k) {
    x = log2(n);
    f = 0;
    for (long long i = x; i >= 0; i--) {
      long long y = pow(2, i);
      if (n - y < k - 1) {
        f = 1;
        continue;
      } else {
        f = 0;
        n -= y;
        k--;
        v.push_back(y);
        break;
      }
    }
    if (f || (n > 0 && k == 0)) {
      cout << "NO"
           << "\n";
      return 0;
    }
  }
  cout << "YES"
       << "\n";
  for (long long i = 0; i < v.size(); i++) cout << v[i] << " ";
  cout << "\n";
  return 0;
}