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;
vector<long long> v1;
long long cnt[35] = {0}, lol = 0;
long long SetBits(long long n) {
  unsigned long long count = 0;
  while (n) {
    n &= (n - 1);
    count++;
  }
  return count;
}
void Binary(long long n) {
  for (int i = 34; i >= 0; i--) {
    int k = n >> i;
    if (k & 1) {
      cnt[i]++;
      v1.push_back(pow(2, i));
      lol++;
    }
  }
}
int main() {
  long long n, k;
  cin >> n >> k;
  bool poss = true;
  if (k > n) {
    poss = false;
  }
  long long tot = SetBits(n);
  if (k < tot) {
    poss = false;
  }
  if (!poss) {
    cout << "NO";
  } else {
    cout << "YES"
         << "\n";
    Binary(n);
    for (long long i1 = tot; i1 < k; i1++) {
      for (int i = 34; i >= 0; i--) {
        if (cnt[i] != 0) {
          cnt[i - 1] += 2;
          cnt[i] -= 1;
          break;
        }
      }
    }
    for (int i = 0; i < 35; i++) {
      for (long long j = 0; j < cnt[i]; j++) {
        long long bc = pow(2, i);
        cout << bc << " ";
      }
    }
  }
}