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;
int main() {
  vector<int> st;
  st.push_back(1);
  int r = 1;
  int n, k;
  cin >> n >> k;
  int n1 = n;
  vector<int> summ;
  while (r * 2 <= n) {
    st.push_back(r * 2);
    r *= 2;
  }
  reverse(st.begin(), st.end());
  for (int i = 0; i < st.size(); i++) {
    if (n == 0) break;
    if (st[i] <= n) {
      summ.push_back(st[i]);
      n -= st[i];
    }
  }
  if (k > n1) {
    cout << "NO";
    return 0;
  }
  int u = 0;
  while (summ.size() < k && u < summ.size()) {
    if (summ[u] == 1) {
      u++;
      continue;
    }
    summ[u] /= 2;
    summ.push_back(summ[u]);
  }
  if (summ.size() > k) {
    cout << "NO";
    return 0;
  }
  cout << "YES" << endl;
  for (int i = 0; i < summ.size(); i++) {
    cout << summ[i] << " ";
  }
}