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;
template <class c>
struct rge {
  c b, e;
};
template <class c>
rge<c> range(c i, c j) {
  return rge<c>{i, j};
}
template <class c>
auto dud(c* x) -> decltype(cerr << *x, 0);
template <class c>
char dud(...);
struct debug {
  template <class c>
  debug& operator<<(const c&) {
    return *this;
  }
};
using ll = long long;
const int nax = 1e5;
void out(unordered_map<int, int> mp) {
  cout << "YES\n";
  for (auto it : mp) {
    for (int i = 0; i < it.second; i++) {
      cout << it.first << " ";
    }
  }
}
int main() {
  int n, k;
  cin >> n >> k;
  bitset<32> b(n);
  int got = 0;
  multiset<int> uniq;
  unordered_map<int, int> mp;
  for (int i = 31; i >= 0; --i) {
    if (b[i] == 1) {
      int po = pow(2, i);
      uniq.insert(po);
      mp[po]++;
      got++;
    }
  }
  if (got == k) {
    out(mp);
    return 0;
  } else if (got > k) {
    cout << "NO\n";
    return 0;
  }
  int need = k - got;
  while (true) {
    auto it = *uniq.rbegin();
    if (it == 1) {
      cout << "NO\n";
      return 0;
    } else {
      uniq.erase(uniq.lower_bound(it));
      uniq.insert(it / 2);
      uniq.insert(it / 2);
      mp[it]--;
      mp[it / 2] += 2;
      need--;
      if (need == 0) {
        out(mp);
        return 0;
      }
    }
  }
}