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 int mxN = 1e5 + 5;
const long long INF = 1e18;
const long long mod = 1e9 + 7;
void solve() {
  int n, k;
  cin >> n >> k;
  if (k > n || __builtin_popcount(n) > k) {
    cout << "NO";
    return;
  }
  multiset<int> s;
  for (int i = 0; i < 31; i++) {
    if (n & (1 << i)) s.insert(1 << i);
  }
  while (s.size() != k) {
    int x = *s.rbegin();
    s.erase(s.find(x));
    s.insert(x / 2);
    s.insert(x / 2);
  }
  cout << "YES" << '\n';
  for (auto x : s) cout << x << " ";
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int t = 1;
  while (t--) {
    solve();
  }
}