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() {
  int n, k;
  cin >> n;
  cin >> k;
  if (k > n || n == 0) {
    cout << "NO"
         << "\n";
    return 0;
  }
  priority_queue<int> heap;
  int count = 0;
  while (n > 0) {
    if (n % 2 == 1) heap.push(pow(2, count));
    n /= 2;
    count++;
  }
  if (k < heap.size()) {
    cout << "NO"
         << "\n";
    return 0;
  }
  while (heap.size() != k && heap.top() != 1) {
    int temp = heap.top();
    heap.pop();
    heap.push(temp / 2);
    heap.push(temp / 2);
  }
  vector<int> array;
  while (!heap.empty()) {
    array.push_back(heap.top());
    heap.pop();
  }
  cout << "YES"
       << "\n";
  reverse(array.begin(), array.end());
  for (int i = 0; i < array.size(); i++) cout << array[i] << " ";
  cout << "\n";
  return 0;
}