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;
  long long k;
  cin >> n;
  cin >> k;
  int n_copy = n;
  vector<long long> powers;
  long long base = 1;
  while (n_copy != 0) {
    if (n_copy % 2 == 1) powers.push_back(base);
    base *= 2;
    n_copy /= 2;
  }
  if (powers.size() > k) {
    cout << "NO" << endl;
    return 0;
  }
  long long steps = k - powers.size();
  int skip = 0;
  for (int i = 0; i < steps; i++) {
    bool done = false;
    for (int j = skip; j < powers.size(); j++) {
      if (powers[j] != 1) {
        done = true;
        powers.insert(powers.begin() + j + 1, powers[j] / 2);
        powers.insert(powers.begin() + j + 1, powers[j] / 2);
        powers.erase(powers.begin() + j);
        break;
      } else {
        skip = j;
      }
    }
    if (!done) {
      cout << "NO";
      return 0;
    }
  }
  cout << "YES" << endl;
  for (int i = 0; i < powers.size(); i++) {
    cout << powers[i] << " ";
  }
  return 0;
}