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 std::cin;
using std::cout;
using std::map;
using std::min;
using std::string;
using std::vector;
int a[31];
int find_power(int x) {
  int nthpower = 1;
  for (int i = 0; i < 31; i++) {
    nthpower = a[i];
    if (x < nthpower) return a[i - 1];
  }
}
int main() {
  int c = 1;
  for (int i = 0; i < 31; i++) {
    a[i] = c;
    c += c;
  }
  int n, k;
  cin >> n >> k;
  vector<int> v;
  if (n < k) {
    cout << "NO\n";
  } else {
    while (true) {
      if (n <= 0 || k <= 0) break;
      int val = n - k + 1;
      int f = find_power(val);
      v.push_back(f);
      n -= f;
      k--;
    }
    if (n != 0) {
      cout << "NO\n";
    } else {
      cout << "YES\n";
      for (int i = v.size() - 1; i >= 0; i--) {
        cout << v[i] << " ";
      }
      cout << "\n";
    }
  }
  return 0;
}