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<long long> expand(long long n, long long k) {
  vector<long long> res(1 + log2(n));
  vector<long long> result;
  long long i = 0;
  while (n) res[i] = (n & 1) ? pow(2, i) : 0, i++, n >>= 1;
  for (long long j : res)
    if (j > 0) result.push_back(j);
  i = 0;
  while (result.size() != k) {
    if (result[i] > 1)
      result[i] /= 2, result.push_back(result[i]);
    else
      i++;
  }
  return result;
}
int32_t main() {
  long long n, k;
  cin >> n >> k;
  long long m = n, count = 0;
  while (m) {
    count += m & 1;
    m >>= 1;
  }
  if (count > k || k > n) {
    cout << "NO\n";
    return 0;
  }
  cout << "YES\n";
  vector<long long> v = expand(n, k);
  for (long long i = 0; i < v.size(); i++) cout << v[i] << " ";
  return 0;
}