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 mod = 1000000007;
const int MAX = 300009;
long long n, k, m, cnt, j;
multiset<long long> ans;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> n >> k;
  if (n == k) {
    cout << "YES";
    cout << endl;
    for (int i = 0; i < n; i++) cout << 1 << " ";
    return 0;
  }
  m = n;
  while (m) {
    if (m % 2 == 1) ans.insert(pow(2, j));
    j++;
    m /= 2;
  }
  if (ans.size() > k) {
    cout << "NO";
    return 0;
  }
  if (ans.size() == k) {
    cout << "YES";
    cout << endl;
    for (auto i : ans) cout << i << " ";
    return 0;
  }
  while (ans.size() != k) {
    auto t = ans.end();
    t--;
    if (*t == 1) break;
    long long tt = *t;
    ans.erase(t);
    ans.insert((tt) / 2);
    ans.insert((tt) / 2);
  }
  if (ans.size() == k) {
    cout << "YES";
    cout << endl;
    for (auto i : ans) cout << i << " ";
    return 0;
  }
  cout << "NO";
}