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;
long long int func(long long int x, long long int a[], long long int n) {
  long long int i = 0, j = n - 1, c = 0;
  while (i < j) {
    if ((a[i] + a[j]) == x) {
      c++, i++, j--;
    } else if ((a[i] + a[j]) < x)
      i++;
    else
      j--;
  }
  return c;
}
void solve() {
  long long int n, k, i;
  cin >> n >> k;
  stack<long long int> q;
  long long int cnt = 0;
  if (n % 2 == 1) {
    cnt++;
  }
  long long int on = cnt;
  for (i = 1; i < 30; i++) {
    if (n & (1ll << i)) q.push(i);
  }
  cnt += q.size();
  if (k < cnt || k > n) {
    cout << "NO" << endl;
    return;
  }
  cout << "YES" << endl;
  while (cnt < k) {
    if (q.top() == 1) {
      cnt++;
      on += 2;
      q.pop();
    } else {
      cnt++;
      long long int h = q.top();
      q.pop();
      q.push(h - 1);
      q.push(h - 1);
    }
  }
  while (on--) {
    cout << 1 << " ";
  }
  while (!q.empty()) {
    long long int y = q.top();
    cout << (1ll << y) << " ";
    q.pop();
  }
  cout << endl;
}
int main() {
  long long int t;
  t = 1;
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  while (t--) {
    solve();
  }
  return 0;
}