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 power(long long n) {
  long long prod = 1;
  for (long long i = 1; i <= n; i++) prod *= 2;
  return prod;
}
int main() {
  long long n, k, cnt = 0, cnt1 = 0, sum = 0;
  cin >> n >> k;
  queue<long long> q;
  long long temp = n;
  while (n) {
    if (n & 1) {
      q.push(power(cnt1));
      cnt++;
    }
    cnt1++;
    n = n >> 1;
  }
  if (k < cnt) {
    cout << "NO\n";
  } else if (k > temp) {
    cout << "NO\n";
  } else {
    cout << "YES\n";
    while (k != cnt) {
      long long num = q.front();
      q.pop();
      if (num == 1)
        q.push(num);
      else {
        q.push(num / 2);
        q.push(num / 2);
        cnt++;
      }
    }
    while (!q.empty()) {
      cout << q.front() << " ";
      q.pop();
    }
  }
  return 0;
}