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 n, k;
  cin >> n >> k;
  if (k > n) {
    cout << "NO" << endl;
    return 0;
  }
  if (k == n) {
    cout << "YES" << endl;
    for (int i = 0; i < n; i++) {
      cout << 1 << " ";
    }
    cout << endl;
    return 0;
  }
  vector<int> powers;
  for (int i = 1, p = 0; i <= n; i *= 2, p++) {
    powers.push_back(i);
  }
  vector<int> result;
  for (int i = powers.size() - 1; i >= 0; i--) {
    if (n == 0) break;
    if (n >= powers[i]) {
      n -= powers[i];
      result.push_back(powers[i]);
      k--;
    }
  }
  if (k < 0) {
    cout << "NO" << endl;
    return 0;
  }
  for (int i = 0; i < result.size(); i++) {
    while (result[i] != 1) {
      if (k == 0) {
        cout << "YES" << endl;
        sort(result.begin(), result.begin() + result.size());
        for (int i = 0; i < result.size(); i++) {
          cout << result[i] << " ";
        }
        cout << endl;
        return 0;
      } else {
        int n = result[i] / 2;
        result.erase(result.begin() + i);
        result.push_back(n);
        result.push_back(n);
        k--;
      }
    }
  }
}