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 const mod = pow(2, 32);
inline long long add(long long a, long long b) {
  return ((a % mod) + (b % mod) + mod) % mod;
}
inline long long mul(long long a, long long b) {
  return (((a % mod) * (b % mod)) % mod + mod) % mod;
}
inline long long sub(long long a, long long b) {
  return ((a % mod) - (b % mod) + mod) % mod;
}
long long power(long long x, long long y, long long m) {
  if (y == 0) return 1;
  long long p = power(x, y / 2, m) % m;
  p = (p * p) % m;
  return (y % 2 == 0) ? p : (x * p) % m;
}
long long modInv(long long x) { return power(x, mod - 2, mod) % mod; }
int main() {
  ios::sync_with_stdio(0);
  cin.tie(NULL);
  cout.tie(NULL);
  long long n, k;
  cin >> n >> k;
  long long x = n;
  long long d = 1;
  vector<long long> v;
  priority_queue<long long> pq;
  while (x) {
    long long z = x % 2;
    if (z) {
      v.push_back(d);
      pq.push(d);
    }
    d = d * 2;
    x = x / 2;
  }
  long long sz = v.size();
  if (sz > k) {
    cout << "NO\n";
  } else {
    while (sz < k) {
      long long x = pq.top();
      pq.pop();
      if (x != 1) {
        pq.push(x / 2);
        pq.push(x / 2);
        sz++;
      } else {
        break;
      }
    }
    if (sz == k) {
      cout << "YES\n";
      while (pq.empty() == 0) {
        cout << pq.top() << " ";
        pq.pop();
      }
    } else {
      cout << "NO\n";
    }
  }
  return 0;
}