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;
const int maxn = 1e5 + 5;
long long a[maxn];
bool cmp(int x, int y) { return x < y; }
priority_queue<long long, vector<long long>, less<long long> > q;
inline long long check(long long x) {
  long long i;
  for (i = 0; i <= 33; i++) {
    if ((1LL << i) > x) break;
  }
  return 1LL << (i - 1);
}
int main() {
  long long n, k;
  cin >> n >> k;
  if (k > n) {
    cout << "NO" << endl;
    return 0;
  }
  long long temp = n;
  while (temp) {
    q.push(check(temp));
    temp -= check(temp);
    k--;
  }
  if (k < 0) {
    cout << "NO" << endl;
    return 0;
  }
  cout << "YES" << endl;
  while (k) {
    long long p = q.top();
    q.pop();
    if (p != 1) {
      long long x = p >> 1;
      q.push(x);
      q.push(x);
      k--;
    }
  }
  while (!q.empty()) {
    cout << q.top() << " ";
    q.pop();
  }
  return 0;
}