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;
multiset<int, greater<int> > st;
int main() {
  int n, k;
  cin >> n >> k;
  int mn = __builtin_popcount(n);
  if (mn > k) return cout << "NO\n", 0;
  int mx = 0;
  for (int i = 0; i <= 30; i++)
    if (n & (1 << i)) st.insert(1 << i), mx += (1 << (i + 1)) - 1;
  if (mx < k) return cout << "NO\n", 0;
  while (st.size() < k) {
    int val = 0;
    for (auto x : st) {
      if (x > 1) {
        val = x;
        break;
      }
    }
    if (val)
      st.erase(st.find(val)), st.insert(val / 2), st.insert(val / 2);
    else
      break;
  }
  if (st.size() != k) return cout << "NO\n", 0;
  cout << "YES\n";
  for (auto x : st) cout << x << " ";
}