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;
using ll = long long int;
int main() {
  ll n, k;
  cin >> n >> k;
  vector<ll> st;
  for (ll i = 0; i < 31; i++) {
    st.push_back(pow(2, i));
  }
  vector<ll> tmp;
  ll t = k;
  ll ptr = 30;
  priority_queue<int, vector<int>, greater<int> > q;
  while (n > 0 && ptr >= 0) {
    if (st[ptr] <= n) {
      n -= st[ptr];
      q.push(st[ptr]);
      k--;
    } else
      ptr--;
  }
  if (tmp.size() > k) {
    cout << "NO";
    return 0;
  }
  vector<ll> answer;
  while (k > 0 && !q.empty()) {
    ll t = q.top();
    q.pop();
    if (t >= 2) {
      q.push(t / 2);
      q.push(t / 2);
      k--;
    } else {
      answer.push_back(t);
    }
  }
  while (!q.empty()) {
    answer.push_back(q.top());
    q.pop();
  }
  if (answer.size() != t) {
    cout << "NO";
    return 0;
  }
  cout << "YES\n";
  reverse(answer.begin(), answer.end());
  for (int i = 0; i < answer.size(); ++i) {
    cout << answer[i] << " ";
  }
  return 0;
}