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;
using ll = long long;
ll mod = 1e9 + 7;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  ll t, n, m, i, j, k;
  cin >> n >> k;
  vector<ll> v;
  i = 0;
  if (n % 2 == 1) {
    v.push_back(1);
    n -= 1;
    i++;
  }
  if (n > 0) {
    j = log2(n);
    m = pow(2, j);
    while (n > 0) {
      if (m <= n) {
        n -= m;
        v.push_back(m);
        i++;
      } else {
        j--;
        m = pow(2, j);
      }
    }
  }
  ll x = 0, y;
  if (i <= k && n == 0) {
    for (j = 0; j < v.size(); j++) {
      if (i == k) break;
      m = v[j];
      if (m == 1) continue;
      v.erase(v.begin() + j);
      j--;
      while (m != 1) {
        m /= 2;
        v.push_back(m);
        i++;
        if (i == k) break;
      }
      v.push_back(m);
    }
    if (i == k) {
      sort(v.begin(), v.end());
      cout << "YES"
           << "\n";
      for (i = 0; i < v.size(); i++) cout << v[i] << " ";
    } else
      cout << "NO";
  } else
    cout << "NO";
}