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 print(int num, int max) {
  if (num == 1) {
    cout << 1 << ' ';
    return 1;
  }
  if (max == 1) {
    cout << num << ' ';
    return 1;
  }
  int cnt = print(num / 2, max - 1);
  if (max > cnt) cnt += print(num / 2, max - cnt);
  return cnt;
}
int main(int argc, char *argv[]) {
  cin.tie(0);
  ios_base::sync_with_stdio(false);
  long long n, k;
  cin >> n >> k;
  if (k > n) {
    cout << "NO" << '\n';
    return 0;
  }
  vector<long long> stack;
  long long idx = 0LL;
  while (n / (1LL << idx)) ++idx;
  --idx;
  while (n) {
    if (n >= (1LL << idx)) {
      stack.push_back(1LL << idx);
      n -= (1LL << idx);
    } else {
      --idx;
    }
  }
  if (stack.size() > k) {
    cout << "NO" << '\n';
    return 0;
  } else {
    cout << "YES" << '\n';
    int lack = k - stack.size();
    for (int i = stack.size() - 1; i >= 0; --i) {
      lack -= (print(stack[i], lack + 1) - 1);
    }
    cout << '\n';
  }
  return 0;
}