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> v;
signed main() {
  long long n, k;
  cin >> n >> k;
  v.push_back(1);
  long long id = 1;
  for (long long i = 1; i <= 32; ++i) {
    id *= 2;
    v.push_back(id);
  }
  long long d = n;
  vector<long long> sum;
  for (long long i = v.size() - 1; i >= 0; --i) {
    while (d >= v[i]) {
      d -= v[i];
      sum.push_back(v[i]);
    }
  }
  if (sum.size() > k || k > n) {
    cout << "NO" << endl;
    return 0;
  }
  cout << "YES" << endl;
  long long sz = sum.size();
  long long i = 0;
  while (sz < k) {
    if (sum[i] > 1) {
      ++sz;
      sum.push_back(sum[i] / 2);
      sum.push_back(sum[i] / 2);
    } else
      sum.push_back(sum[i]);
    ++i;
  }
  for (long long i = sum.size() - 1; i >= sum.size() - k; --i) {
    cout << sum[i] << " ";
  }
  return 0;
}