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;
constexpr int INF = numeric_limits<int>::max();
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int n, k;
  cin >> n >> k;
  if (n < k || k < __builtin_popcount(n)) {
    cout << "NO";
    exit(0);
  }
  cout << "YES\n";
  stack<int> num;
  int x = 1;
  int sz = 0;
  while (n >= x) {
    if (n & x) {
      num.push(x);
      sz++;
    }
    x <<= 1;
  }
  while (sz < k) {
    int t = num.top();
    num.pop();
    if (t == 1) {
      cout << "1 ";
    } else {
      num.push(t >> 1);
      num.push(t >> 1);
      sz++;
    }
  }
  while (!num.empty()) {
    cout << num.top() << " ";
    num.pop();
  }
}