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 n, k, a[200006], i, j, ans, f[33], cnt = 0, x;
int main() {
  f[0] = 1;
  for (i = 1; i <= 32; i++) f[i] = f[i - 1] * 2;
  scanf("%I64d%I64d", &n, &k);
  if (k > n)
    printf("NO");
  else {
    if (k == 1) {
      for (j = 0; j <= 31; j++) {
        if (f[j] >= n) break;
      }
      if (n == f[j])
        printf("YES\n%I64d", n);
      else
        printf("NO");
    } else {
      if (n % 2 == 1) {
        cnt++;
        a[0] = 1;
        k -= 1;
        n--;
      }
      ans = n;
      for (i = 1; i < k; i++) {
        ans -= k - i;
        for (j = 0; j <= 31; j++) {
          if (f[j] <= ans && f[j + 1] > ans) break;
        }
        a[i] = f[j];
        ans = ans + k - i - a[i];
      }
      for (j = 0; j <= 31; j++) {
        if (f[j] <= ans && f[j + 1] > ans) break;
      }
      if (ans == f[j]) {
        printf("YES\n");
        a[k] = ans;
        if (cnt == 1) printf("%d ", a[0]);
        for (i = 1; i <= k; i++) printf("%I64d ", a[i]);
      } else
        printf("NO");
    }
  }
}