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 mult(long long a, long long b, long long p = 1000000007) {
  return ((a % p) * (b % p)) % p;
}
long long add(long long a, long long b, long long p = 1000000007) {
  return (a % p + b % p) % p;
}
long long fpow(long long n, long long k, long long p = 1000000007) {
  long long r = 1;
  for (; k; k >>= 1LL) {
    if (k & 1LL) r = mult(r, n, p);
    n = mult(n, n, p);
  }
  return r;
}
long long inv(long long a, long long p = 1000000007) {
  return fpow(a, p - 2, p);
}
void self_max(int &a, int &b) { a = max(a, b); }
void self_min(int &a, int &b) { a = min(a, b); }
const int INF = 1e9 + 5;
void solve() {
  multiset<long long, greater<long long> > s;
  long long n, k;
  cin >> n >> k;
  int cnt = __builtin_popcount(n);
  if (k > n || k < cnt) {
    cout << "NO" << '\n';
    return;
  }
  for (int i = 0; i < 64; i++) {
    if (n >> i & 1) {
      s.insert(pow(2, i));
    }
  }
  while (s.size() != k) {
    long long no = *s.begin();
    cerr << "no"
         << ": " << no << '\n';
    s.erase(s.begin());
    s.insert(no / 2);
    s.insert(no / 2);
  }
  cout << "YES" << '\n';
  for (int i : s) cout << i << " ";
  cout << '\n';
}
int main() {
  ios::sync_with_stdio(false);
  cout << fixed << setprecision(20);
  cin.tie(0);
  cout.tie(0);
  int t = 1;
  while (t--) {
    solve();
  }
}