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 long double pi = 3.141592653;
const long long mod = 1e9 + 7;
const long long N = 5e5 + 500;
const int big = 1e9 + 1;
const long long INF = 1e18 + 1;
long long n, k, ans, pw = 2;
string s;
queue<long long> q;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> k;
  long long lol = n;
  while (n) {
    if (n % 2) {
      ans++;
      s += '1';
    } else
      s += '0';
    n /= 2;
  }
  if (ans > k || lol < k) {
    cout << "NO";
    return 0;
  }
  for (int i = 1; i < s.size(); ++i) {
    pw *= 2;
  }
  n = lol;
  for (int i = s.size(); i > 0; --i) {
    if (s[i] == '1') {
      q.push(pw);
    }
    pw /= 2;
  }
  if (n % 2) {
    while (q.size() < k - 1) {
      if (q.front() == 1) {
        q.push(q.front());
        q.pop();
      } else {
        long long cur = q.front() / 2;
        if (cur >= 1) {
          q.push(cur);
          q.push(cur);
        }
        q.pop();
      }
    }
    q.push(1);
  } else {
    while (q.size() < k) {
      if (q.front() == 1) {
        q.push(q.front());
        q.pop();
      } else {
        long long cur = q.front() / 2;
        if (cur >= 1) {
          q.push(cur);
          q.push(cur);
        }
        q.pop();
      }
    }
  }
  cout << "YES\n";
  while (!q.empty()) {
    cout << q.front() << " ";
    q.pop();
  }
  return 0;
}