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 int INF = 1e18;
long long int MOD = 1e9 + 7;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long int i, j, k, l, m, n, p, p1, q, r, cnt = 0, maxi = 0, mini = INF, a,
                                               b, c, d, ans, t, x, x1;
  cin >> n >> k;
  l = n;
  j = 0;
  long long int C[32];
  for (i = 0; i < 32; i++) {
    C[i] = 0;
  }
  while (l > 0) {
    if (l % 2 == 1) {
      cnt++;
      C[j]++;
    }
    l = l / 2;
    j++;
  }
  if (cnt > k || k > n) {
    cout << "NO" << endl;
  } else {
    k = k - cnt;
    cout << "YES" << endl;
    while (k > 0) {
      for (i = 31; i >= 1; i--) {
        if (C[i] > 0) {
          C[i]--;
          C[i - 1] = C[i - 1] + 2;
          k--;
          break;
        }
      }
    }
    for (i = 0; i < 32; i++) {
      x = pow(2, i);
      for (j = 0; j < C[i]; j++) {
        cout << x << " ";
      }
    }
  }
}