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;
signed main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  long long n, k;
  cin >> n >> k;
  vector<long long> powers(63, 0);
  long long tot = 0;
  for (long long i = 0; i < 63; i++) {
    if ((n >> i) & 1) {
      powers[i]++;
      tot++;
    }
  }
  if (tot > k) {
    cout << "NO" << endl;
    return 0;
  }
  long long left = k - tot;
  for (long long i = 62; i > 0; i--) {
    if (left > powers[i]) {
      left -= powers[i];
      powers[i - 1] += 2 * powers[i];
      powers[i] = 0;
    } else {
      powers[i] -= left;
      powers[i - 1] += 2 * left;
      left = 0;
      break;
    }
  }
  if (left != 0) {
    cout << "NO" << endl;
    return 0;
  }
  cout << "YES" << endl;
  for (long long i = 0; i < 63; i++) {
    for (long long j = 0; j < powers[i]; j++) {
      cout << (long long)pow(2, i) << ' ';
    }
  }
  cout << endl;
}