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;
const long long mod = 998244353;
inline void solve() {
  long long n, k;
  cin >> n >> k;
  priority_queue<long long> pq;
  while (n > 0) {
    long long pre = n;
    for (long long i = 0, p = 1; i <= 31; i++, p *= 2) {
      if (p > n) {
        pq.push(p / 2);
        n -= p / 2;
        break;
      }
    }
    if (pre == n) break;
  }
  while (pq.size() < k and pq.top() % 2 == 0) {
    long long x = pq.top();
    pq.pop();
    pq.push(x / 2);
    pq.push(x / 2);
  }
  if ((long long)pq.size() == k) {
    cout << "YES" << '\n';
    while (!pq.empty()) {
      cout << pq.top() << ' ';
      pq.pop();
    }
  } else
    cout << "NO" << endl;
}
signed main() {
  ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  long long n = 1;
  while (n--) solve();
  return 0;
}