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 findMaxPow(int n) {
  int i = 1;
  while (n > i) {
    i *= 2;
  }
  if (n < i) {
    i /= 2;
  }
  return i;
}
void isPossible(int n, int k, vector<int> &v) {
  int remaining = n, maxPow;
  do {
    maxPow = findMaxPow(remaining);
    remaining -= maxPow;
    v.push_back(maxPow);
  } while (remaining > 0);
  if (v.size() >= k) {
    return;
  }
  int i = 0;
  while (v.size() < k && i < v.size()) {
    if (v[i] == 1) {
      i++;
      continue;
    }
    int curr = v[i];
    v[i] /= 2;
    v.push_back(curr / 2);
  }
}
int main() {
  int n, k;
  vector<int> sol;
  cin >> n >> k;
  isPossible(n, k, sol);
  if (sol.size() == k) {
    cout << "YES" << endl;
    for (auto x : sol) {
      cout << x << ' ';
    }
  } else {
    cout << "NO";
  }
}