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 int MAXN = 200100;
const int INF = 2000000100;
int N, K;
int pnum[35];
void output() {
  cout << "YES\n";
  bool s = false;
  for (int i = 0; i < 32; i++) {
    for (int j = 0; j < pnum[i]; j++) {
      if (s) cout << " ";
      s = true;
      cout << (1 << i);
    }
  }
  cout << "\n";
}
int main() {
  ios_base::sync_with_stdio(0);
  cin >> N >> K;
  if (K > N) {
    cout << "NO\n";
    return 0;
  }
  pnum[0] = N;
  if (K == N) {
    output();
    return 0;
  }
  int k = N;
  for (int i = 0; i < 32; i++) {
    if (pnum[i] == 0) {
      cout << "NO\n";
      return 0;
    }
    if (k - pnum[i] / 2 > K) {
      k -= pnum[i] / 2;
      pnum[i + 1] = pnum[i] / 2;
      pnum[i] -= 2 * (pnum[i] / 2);
    } else {
      pnum[i + 1] = k - K;
      pnum[i] -= 2 * (k - K);
      output();
      return 0;
    }
  }
  cout << "NO\n";
}