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[32], x[32];
bool judge(int n) {
  for (int i = 0; i <= 31; i++)
    if (n == x[i]) return true;
  return false;
}
int main() {
  int n, k, u = 1;
  scanf("%d%d", &n, &k);
  for (int j = 0; j < 32; j++) x[j] = u, u *= 2;
  if (k > n || (k == 1 && !judge(n))) {
    printf("NO\n");
    return 0;
  }
  int i = 0;
  while (n) {
    a[i] = n % 2;
    n /= 2;
    i++;
  }
  int cnt = 0;
  for (int j = 0; j < 32; j++)
    if (a[j]) cnt++;
  if (cnt > k) {
    printf("NO\n");
    return 0;
  }
  while (cnt < k)
    for (int j = 31; j >= 1; j--) {
      if (cnt == k) break;
      if (a[j]) {
        for (int h = 0; h < j; h++) {
          if (k - cnt >= x[j - h] - 1) {
            cnt += x[j - h] - 1;
            a[h] += x[j - h];
            a[j]--;
            break;
          }
        }
      }
    }
  printf("YES\n");
  for (int j = 0; j <= 31; j++)
    for (int h = 0; h < a[j]; h++) printf("%d ", x[j]);
  return 0;
}