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 n, k, a[50], cnt, tot, mi[35];
int main() {
  scanf("%d%d", &n, &k);
  mi[0] = 1;
  for (int i = 1; i <= 30; i++) mi[i] = mi[i - 1] << 1;
  while (n) {
    a[tot++] = n % 2;
    if (n % 2) cnt++;
    n /= 2;
  }
  if (k < cnt) {
    printf("NO");
    return 0;
  }
  for (int i = tot - 1; i > 0; i--) {
    if (cnt == k) break;
    while (a[i]) {
      cnt++;
      a[i - 1] += 2;
      a[i]--;
      if (cnt == k) break;
    }
  }
  if (k != cnt) {
    printf("NO");
    return 0;
  }
  printf("YES\n");
  for (int i = 0; i < tot; i++) {
    for (int j = 0; j < a[i]; j++) {
      printf("%d ", mi[i]);
    }
  }
  return 0;
}