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_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long int t;
  long long int n, k;
  cin >> n >> k;
  if (n < k) {
    cout << "NO\n";
    return 0;
  }
  vector<long long int> a(32, 0);
  long long int i = 0;
  long long int cnt = 0;
  while (n) {
    if (n & 1) {
      a[i]++;
      cnt++;
    }
    i++;
    n /= 2;
  }
  reverse(a.begin(), a.end());
  for (long long int i = 0; i < 31; ++i) {
    if (a[i]) {
      while (a[i] && cnt < k) {
        a[i]--;
        a[i + 1] += 2;
        cnt++;
      }
    }
  }
  reverse(a.begin(), a.end());
  if (cnt != k) {
    cout << "NO\n";
  } else {
    cout << "YES\n";
    for (long long int i = 0; i < 32; ++i) {
      if (a[i]) {
        while (a[i]--) {
          cout << (1 << i) << " ";
        }
      }
    }
  }
}