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;
long long int power[100];
string binary;
void binary_convert(long long int a) {
  while (a != 0) {
    if (a % 2 == 0) {
      binary += "0";
    } else {
      binary += "1";
    }
    a /= 2;
  }
}
int main() {
  long long int i, a;
  power[0] = 1;
  for (i = 1; i <= 32; i++) {
    power[i] = power[i - 1] * 2;
  }
  long long int n, k;
  priority_queue<long long int> ans;
  cin >> n >> k;
  if (n < k) {
    cout << "NO" << endl;
    return 0;
  }
  binary_convert(n);
  for (i = 0; i < binary.size(); i++) {
    if (binary[i] == '1') {
      ans.push(power[i]);
    }
  }
  if (k < ans.size()) {
    cout << "NO" << endl;
    return 0;
  }
  while (k != ans.size()) {
    a = ans.top();
    ans.pop();
    a /= 2;
    ans.push(a);
    ans.push(a);
  }
  cout << "YES" << endl;
  while (!ans.empty()) {
    cout << ans.top() << " ";
    ans.pop();
  }
}