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;
typedef long long ll;
void showpq(vector<ll> gq) {
  vector<ll> g = gq;
  reverse(g.begin(), g.end());
  while (!g.empty()) {
    cout << g.back() << " ";
    g.pop_back();
  }
  cout << '\n';
}
bool mai(string a, string b) { return a.size() > b.size(); }
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  ll n, k, chk = 1, a, f = 0;
  vector<ll> v;
  cin >> n >> k;
  if (n < k)
    cout << "NO";
  else {
    a = n;
    while (a != 0) {
      if (chk > k) {
        f = 1;
        break;
      }
      v.push_back(pow(2, ((ll)(log2(double(a))))));
      a = a - pow(2, ((ll)(log2(double(a)))));
      chk++;
    }
    chk--;
    if (f == 1) {
      cout << "NO";
    } else {
      for (ll i = 0, j = 0; i < (k - chk); ++i) {
        if (v[j] != 1) {
          v[j] = (ll)(v[j] / 2);
          v.push_back(v[j]);
        } else {
          --i;
          j++;
        }
      }
      cout << "YES" << endl;
      sort(v.begin(), v.end());
      showpq(v);
    }
  }
  return 0;
}