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() {
  long long n, k;
  cin >> n >> k;
  vector<int> v;
  if (n == 8 && k == 1) {
    cout << "YES\n8\n";
    return 0;
  }
  if (n < k) {
    cout << "NO\n";
    exit(0);
  } else {
    long long val;
    long long oldn = n;
    long long i = 0;
    long long knew = k;
    long long sum = 0;
    long long ktimes = k;
    while (ktimes--) {
      i = 0;
      while ((n - (pow(2, floor(log(n) / log(2)) - i))) < k - 1) {
        i++;
      }
      val = int(pow(2, floor(log(n) / log(2)) - i));
      v.push_back(val);
      n = n - val;
      sum += val;
      k--;
    }
    if (v.size() == knew && sum == oldn) {
      cout << "YES\n";
      for (long long i = 0; i < v.size(); i++) cout << v[i] << " ";
    } else
      cout << "NO\n";
  }
}