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 long long INF = LLONG_MAX;
const long long base = 1e9;
const double PI = acos(-1);
const long long MOD = 1e9 + 7;
const long long N = 1e5 + 10;
int main() {
  ios_base ::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long n, k;
  cin >> n >> k;
  long long a[k];
  if (n < k) {
    cout << "NO"
         << "\n";
    return 0;
  }
  a[0] = n - k + 1;
  for (int i = 1; i < k; i++) {
    a[i] = 1;
  }
  for (int i = 0; i < k; i++) {
    long long y = 0;
    while (1) {
      if (pow(2, y) == a[i]) {
        cout << "YES"
             << "\n";
        for (int j = 0; j < k; j++) {
          cout << a[j] << " ";
        }
        cout << endl;
        return 0;
      }
      if (pow(2, y) > a[i]) {
        a[i + 1] += a[i] - pow(2, y - 1);
        a[i] = pow(2, y - 1);
        break;
      }
      y++;
    }
  }
  cout << "NO"
       << "\n";
  return 0;
}