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() {
  int tc = 1;
  while (tc--) {
    ios_base::sync_with_stdio(false);
    long long n, k, cnt = 0, i, x = 1;
    cin >> n >> k;
    vector<int> v;
    if (k <= n) {
      while (n) {
        v.push_back(n % 2);
        cnt += n % 2;
        n /= 2;
      }
      for (i = v.size() - 1; i >= 0; i--) {
        if (cnt >= k) break;
        while (v[i]) {
          if (cnt >= k)
            break;
          else {
            v[i]--;
            cnt++;
            v[i - 1] += 2;
          }
        }
      }
      if (cnt == k) {
        cout << "YES" << endl;
        for (i = 0; i < v.size(); i++) {
          x = pow(2, i);
          while (v[i]) {
            cout << x << " ";
            v[i]--;
          }
        }
      } else
        cout << "NO" << endl;
    } else
      cout << "NO" << endl;
  }
  return 0;
}