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;
long long countOnes(long long n) {
  long long c = 0;
  while (n > 0) {
    if (n % 2 != 0) {
      c++;
    }
    n = n / 2;
  }
  return c;
}
void coutOnes(long long n, long long k) {
  vector<long long> v;
  long long c = countOnes(n);
  long long po2 = 1;
  while (n > 0) {
    if (n % 2 != 0) {
      v.push_back(po2);
    }
    po2 *= 2;
    n = n / 2;
  }
  for (int i = c - 1; i >= 0; i--) {
    if (v[i] <= k - i) {
      k -= v[i];
      for (int j = 0; j < v[i]; j++) cout << 1 << " ";
    } else {
      long long kl = v[i];
      while (v[i] != 0) {
        if (v[i] / kl > k - i) {
          kl *= 2;
          for (int j = 0; j < k - i - v[i] / kl; j++)
            cout << kl / 2 << " " << kl / 2 << " ";
          for (int j = k - i - v[i] / kl; j < v[i] / kl; j++) cout << kl << " ";
          for (int j = 0; j < i; j++) cout << v[j] << " ";
          return;
        }
        kl /= 2;
      }
    }
  }
}
int main() {
  long long n, k;
  cin >> n >> k;
  int i;
  bool f = true;
  if (n < k || countOnes(n) > k) {
    return cout << "NO", 0;
  }
  cout << "YES" << endl;
  coutOnes(n, k);
}