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 int __xor(long long int a, long long int b) { return a ^ b; }
long long int __and(long long int a, long long int b) { return a & b; }
long long int __or(long long int a, long long int b) { return a | b; }
long long int ai[1000], pi[1000];
int main() {
  long long int b, c, y, z, t, i, j, k = 1, l1, l2, m = 0, n, c1 = 0, c2 = 0,
                                     f = 0, x;
  scanf("%lld", &n);
  scanf("%lld", &y);
  l1 = n;
  vector<long long int> vec;
  queue<long long int> q;
  while (l1 != 0) {
    l2 = l1 % 2;
    l1 = l1 / 2;
    if (l2 == 1) {
      q.push(k);
    }
    k = k * 2;
  }
  if (q.size() > y)
    cout << "NO";
  else if (q.size() == y) {
    cout << "YES\n";
    while (!q.empty()) {
      cout << q.front() << " ";
      q.pop();
    }
  } else {
    while (!q.empty()) {
      if (q.front() > 1) {
        l1 = q.front();
        q.pop();
        q.push(l1 / 2);
        q.push(l1 / 2);
        if (q.size() + vec.size() == y) {
          f = 1;
          break;
        }
      } else {
        vec.push_back(1);
        q.pop();
        if (q.size() + vec.size() == y) {
          f = 1;
          break;
        }
      }
    }
    if (f == 1) {
      cout << "YES\n";
      for (j = 0; j < vec.size(); j++) cout << vec[j] << " ";
      while (!q.empty()) {
        cout << q.front() << " ";
        q.pop();
      }
    } else
      cout << "NO";
  }
  return 0;
}