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 gcd(int a, int b) { return b ? gcd(b, a % b) : a; }
int main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  int n, k, i, cnt = 0, tmp;
  scanf("%d%d", &n, &k);
  vector<int> a;
  tmp = n;
  for (i = 0; tmp > 0; i++) {
    if (((n) >> (i)&1)) {
      a.push_back(((n) >> (i)&1) * pow(2, i));
      cnt++;
    }
    tmp >>= 1;
  }
  if (k < cnt || k > n)
    printf("NO\n");
  else {
    printf("YES\n");
    while (cnt < k) {
      int z;
      z = a.back();
      a.pop_back();
      if (z / 2 > 1) {
        a.push_back(z / 2);
        a.push_back(z / 2);
      }
      cnt++;
    }
    for (int i = 0; i < a.size(); i++) printf("%d ", a[i]);
    for (int i = 0; i < k - a.size(); i++) printf("1 ");
  }
  return 0;
}