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, cur;
priority_queue<int> pq;
int main() {
  scanf("%d %d", &n, &k);
  cur = 1;
  while (n) {
    if (n % 2 == 1) {
      pq.push(cur);
    }
    n /= 2;
    cur *= 2;
  }
  if (pq.size() > k)
    printf("NO\n");
  else {
    while (k > pq.size()) {
      int now = pq.top();
      if (now == 1) {
        printf("NO\n");
        return 0;
      }
      pq.pop();
      pq.push(now / 2);
      pq.push(now / 2);
    }
    printf("YES\n");
    while (!pq.empty()) {
      int now = pq.top();
      printf("%d", now);
      pq.pop();
      if (pq.empty())
        printf("\n");
      else
        printf(" ");
    }
  }
}