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 double pi = acos(-1.0);
const double pii = 2 * pi;
const double eps = 1e-6;
int main() {
  std::ios::sync_with_stdio(false);
  std::cin.tie(NULL);
  std::cout.tie(NULL);
  int n, k;
  std::cin >> n >> k;
  std::vector<int> bn{};
  {
    int a, b;
    a = n;
    b = 1;
    while (a) {
      if (a % 2) {
        bn.push_back(b);
      }
      a /= 2;
      b *= 2;
    }
  }
  if (k < bn.size() || k > n) {
    std::cout << "NO\n";
    return 0;
  }
  std::cout << "YES" << std::endl;
  std::queue<int> q{}, ans{};
  for (int i = (int)0; i < (int)bn.size(); i++) {
    q.push(bn[i]);
  }
  while (!q.empty()) {
    if (q.size() == k) {
      while (!q.empty()) {
        ans.push(q.front());
        q.pop();
      }
    } else {
      if (q.front() == 1) {
        q.pop();
        ans.push(1);
        k--;
      } else {
        int a = q.front();
        q.pop();
        q.push(a / 2);
        q.push(a / 2);
      }
    }
  }
  while (!ans.empty()) {
    std::cout << ans.front() << " ";
    ans.pop();
  }
  std::cout << "\n";
  return 0;
}