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 a[100001];
deque<long long int> v;
int main() {
  ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  ;
  long long int n, k;
  cin >> n >> k;
  if (k > n) {
    cout << "NO";
    return 0;
  }
  if (k == n) {
    cout << "YES\n";
    for (long long int i = 0; i < k; i = i + 1) {
      cout << 1 << " ";
    }
    return 0;
  }
  long long int cn = n;
  while (cn > 0) {
    long long int lg = log2(cn);
    long long int f = pow(2, lg);
    cn = cn - f;
    v.push_front(f);
  }
  if (v.size() > k) {
    cout << "NO";
    return 0;
  }
  if (v.size() == k) {
    cout << "YES\n";
    for (long long int i = 0; i < v.size(); i = i + 1) {
      cout << v[i] << " ";
    }
    return 0;
  }
  while (v.size() < k) {
    if (v[v.size() - 1] != 1) {
      v[v.size() - 1] = v[v.size() - 1] / 2;
      v.push_front(v[v.size() - 1]);
      v.push_front(v[v.size() - 1]);
      v.pop_back();
    } else {
      sort(v.begin(), v.end());
      if (v[v.size() - 1] != 1) {
        v[v.size() - 1] = v[v.size() - 1] / 2;
        v.push_front(v[v.size() - 1]);
        v.push_front(v[v.size() - 1]);
        v.pop_back();
      } else {
        cout << "NO";
        return 0;
      }
    }
  }
  sort(v.begin(), v.end());
  if (v.size() == k) {
    cout << "YES\n";
    for (long long int i = 0; i < v.size(); i = i + 1) {
      cout << v[i] << " ";
    }
    return 0;
  } else {
    cout << "NO";
    return 0;
  }
}