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 pow(int a, int n) {
  int pro = 1;
  for (int i = 0; i < n; i++) {
    pro *= 2;
  }
  return pro;
}
int main() {
  long n, k;
  cin >> n >> k;
  if (k > n) {
    cout << "NO" << endl;
    return 0;
  }
  vector<int> v;
  while (n) {
    v.push_back(n % 2);
    n = n / 2;
  }
  for (int i = 0; i < v.size(); i++) {
    k = k - v[i];
  }
  for (int i = v.size() - 1; i >= 0; i--) {
    if (k == 0)
      break;
    else if (k < 0) {
      cout << "NO" << endl;
      return 0;
    } else {
      int h = k - v[i];
      if (h >= 0) {
        k = k - v[i];
        v[i - 1] += 2 * v[i];
        v[i] = 0;
      } else {
        v[i - 1] += 2 * k;
        v[i] -= k;
        k = 0;
      }
    }
  }
  if (k != 0) {
    cout << "NO" << endl;
    return 0;
  }
  cout << "YES" << endl;
  for (int i = 0; i < v.size(); i++) {
    while (v[i]--) {
      cout << pow(2, i) << " ";
    }
  }
  cout << endl;
  return 0;
}