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 F(long long x) {
  long long ret = 0;
  while (x > 0) {
    x /= 2;
    ret++;
  }
  return ret;
}
int main() {
  long long n, k;
  scanf("%lld %lld", &n, &k);
  long long tmp = n;
  long long x = 2 + F(n);
  vector<long long> v;
  for (long long i = x; i >= 0; i--) {
    if ((1LL << i) <= n) {
      n -= (1LL << i);
      v.push_back((1LL << i));
    }
  }
  if (k < v.size() || k > tmp) {
    printf("NO\n");
    return 0;
  }
  printf("YES\n");
  vector<long long> ans;
  k -= ((long long)v.size());
  long long cnt = 0;
  for (long long i = 0; i < v.size(); i++) {
    if (v[i] == 1) {
      printf("%lld ", v[i]);
      continue;
    }
    deque<long long> dq;
    dq.push_back(v[i]);
    while (k) {
      x = dq.front();
      dq.pop_front();
      if (x == 1) {
        dq.push_back(1);
        break;
      }
      dq.push_back(x / 2);
      dq.push_back(x / 2);
      k--;
    }
    while (!dq.empty()) {
      printf("%lld ", (long long)dq.front());
      dq.pop_front();
    }
  }
  printf("\n");
  return 0;
}