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() {
  ios::sync_with_stdio(false);
  cin.tie(), cout.tie();
  long long n, k;
  cin >> n >> k;
  if (k > n) {
    cout << "NO" << endl;
    return 0;
  }
  long long a[k];
  for (int i = 0; i < k; i++) a[i] = 1;
  long long rem = n - k;
  for (int i = 0; i < k; i++) {
    while (2 * a[i] - 1 <= rem) {
      a[i] *= 2;
    }
    rem -= (a[i] - 1);
  }
  long long sum = 0;
  for (int i = 0; i < k; i++) sum += a[i];
  if (sum == n) {
    cout << "YES" << endl;
    for (int i = 0; i < k; i++) cout << a[i] << " ";
    cout << endl;
  } else
    cout << "NO" << endl;
  return 0;
}