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;
const int N = 1e6 + 9;
const long long int mod = 1e9 + 7;
map<long long int, long long int> mpp, mp;
vector<long long int> v;
multiset<long long int> st, stt;
long long int c[N], t[] = {0, 2, 4, 9, 10, 11};
long long int a[N], b[N];
void solve() {
  long long int n, k;
  cin >> n >> k;
  b[0] = 1;
  for (int i = 1; i <= 33; ++i) b[i] = b[i - 1] * 2;
  long long int p = n, i = 1;
  while (p) {
    a[i] = p % 2;
    p /= 2;
    i++;
  }
  i--;
  p = 1;
  for (int j = 1; j <= i; ++j) {
    if (a[j]) {
      c[p] = b[j - 1];
      p++;
    }
  }
  p--;
  sort(c, c + p + 1);
  long long int tag = 1;
  i = p;
  if (p > k) {
    cout << "NO";
    return;
  }
  while (i < k) {
    if (c[tag] > 1) {
      p++;
      c[p] = c[tag] / 2;
      c[tag] /= 2;
      i++;
    } else
      tag++;
    if (c[tag] == 0) break;
  }
  if (i != k) {
    cout << "NO";
    return;
  }
  cout << "YES" << '\n';
  for (int i = 1; i <= p; ++i) cout << c[i] << " ";
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long int t = 1;
  while (t--) solve();
}