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;
int a[60];
int main() {
  int n, k;
  scanf("%d%d", &n, &k);
  int i = 0;
  int t = n;
  while (t) {
    if (t & 1) a[i] = 1;
    i++;
    t >>= 1;
  }
  int l = i;
  int mi = 0, ma = 0;
  for (i = 0; i < l; i++) {
    if (a[i]) {
      mi++;
    }
  }
  if (k > n || k < mi) {
    printf("NO\n");
  } else {
    printf("YES\n");
    int cnt = mi;
    for (i = l - 1; i >= 0; i--) {
      if (cnt < k) {
        int res = k - cnt;
        if (res > a[i]) {
          a[i - 1] += a[i] * 2;
          cnt += a[i];
          a[i] = 0;
        } else {
          a[i - 1] += res * 2;
          a[i] -= res;
          break;
        }
      } else
        break;
    }
    for (i = 0; i < l; i++) {
      for (int j = 0; j < a[i]; j++) {
        printf("%d ", 1 << i);
      }
    }
    printf("\n");
  }
}