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 e = 1000000007;
int gcd(int a, int b) { return b == 0 ? a : gcd(b, a % b); }
void traverse(vector<int> v) {
  for (int i = 0; i < v.size(); ++i) cout << v[i] << " ";
  cout << "\n";
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int i, j, k, l, n, x, y, m, z, a, b, test, temp;
  int total, nos;
  cin >> total >> nos;
  if (total == nos) {
    cout << "YES\n";
    for (int i = 0; i < (total); i++) cout << 1 << " ";
    return 0;
  }
  if (total < nos) {
    cout << "NO\n";
    return 0;
  }
  std::string binary = std::bitset<40>(total).to_string();
  reverse(binary.begin(), binary.end());
  int count = 0;
  priority_queue<int> pq;
  for (int i = 0; i < (40); i++) {
    if (binary[i] == '1') {
      pq.push(pow(2, i));
      count++;
    }
  }
  if (nos < count) {
    cout << "NO";
    return 0;
  }
  cout << "YES\n";
  while (pq.size() != nos) {
    x = pq.top();
    pq.pop();
    pq.push(x / 2);
    pq.push(x / 2);
  }
  for (int i = 0; i < (nos); i++) {
    cout << pq.top() << " ";
    pq.pop();
  }
  return 0;
}