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() {
  long long n, k;
  cin >> n >> k;
  long long temp = n;
  bool not_two = false;
  if (n % 2 != 0) {
    temp -= 1;
    not_two = true;
  }
  long long merge = n - k;
  long long array[100000];
  array[0] = temp;
  long long step = 0;
  bool find = false;
  while (merge > 0) {
    if (array[step] == 0) {
      find = true;
      break;
    }
    if (array[step] >= 2) {
      long long x = min(merge, (array[step] / 2));
      merge -= x;
      array[step] -= x * 2;
      array[step + 1] += x;
    }
    step++;
  }
  if (find == true || merge != 0) {
    if (n != k) {
      cout << "NO" << endl;
      return 0;
    } else {
      cout << "YES" << endl;
      for (int i = 0; i < k; i++) {
        cout << "1 ";
      }
      cout << endl;
      return 0;
    }
  }
  cout << "YES" << endl;
  if (not_two == true) cout << "1 ";
  for (long long i = 0; i < step + 1; i++) {
    if (array[i] == 0) continue;
    for (long long j = 0; j < array[i]; j++) {
      cout << (int)pow(2, i) << " ";
    }
  }
  cout << endl;
}