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 n, k, c;
map<long long, long long> m;
bool check() {
  for (auto i : m) {
    if (i.second < 0) return 0;
  }
  return 1;
}
void solve() {
  cin >> n >> k;
  if (k > n) {
    cout << "NO" << endl;
    return;
  };
  if (k == n) {
    cout << "YES" << endl;
    for (__typeof((n + 1)) i = (1); i < (n + 1); i++) cout << 1 << " ";
    return;
  }
  for (__typeof((62ll)) i = (62ll); i >= (0); i--) {
    if ((n >> i) & 1ll) m[i]++;
  }
  if ((long long)m.size() > k) {
    cout << "NO" << endl;
    return;
  };
  c = (long long)m.size();
  while (c != k) {
    pair<long long, long long> t = *m.rbegin();
    if (t.second + c <= k) {
      m.erase(t.first);
      m[t.first - 1] += 2 * t.second;
      c += t.second;
    } else {
      m[t.first] -= k - c;
      m[t.first - 1] += 2 * (k - c);
      c = k;
    }
  }
  if (!check()) {
    cout << "NO" << endl;
    return;
  };
  cout << "YES" << endl;
  for (auto i : m) {
    long long x = (1ll << i.first);
    for (__typeof((i.second)) j = (0); j < (i.second); j++) cout << x << " ";
  }
}
void prep() {}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long t = 1;
  prep();
  cout << fixed << setprecision(12);
  while (t--) solve();
  return 0;
}