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> dec_to_bin(long long n) {
  vector<long long> bin;
  long long k = 0;
  while (n) {
    long long rem = n % 2;
    bin.push_back(rem);
    n /= 2;
  }
  while (bin.size() < 31) bin.push_back(0);
  return bin;
}
long long pawer(long long n) {
  long long ans = 1;
  long long base = 2;
  while (n) {
    if (n & 1) ans *= base;
    base *= base;
    n >>= 1;
  }
  return ans;
}
void solve() {
  long long n, k;
  cin >> n >> k;
  vector<long long> bin = dec_to_bin(n);
  priority_queue<long long> pq;
  long long p = 0;
  for (long long i = 0; i < bin.size(); i++) {
    if (bin[i]) {
      pq.push(i);
    }
  }
  if (pq.size() > k) {
    cout << "NO\n";
  } else {
    while (pq.size() < k && pq.top()) {
      long long ind = pq.top();
      pq.pop();
      pq.push(ind - 1);
      pq.push(ind - 1);
    }
    if (pq.size() == k) {
      cout << "YES\n";
      while (!pq.empty()) {
        cout << pawer(pq.top()) << ' ';
        pq.pop();
      }
    } else {
      cout << "NO\n";
    }
  }
}
int32_t main() {
  long long t = 1;
  while (t--) solve();
  return 0;
}