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 Int() {
  int x;
  scanf("%d", &x);
  return x;
}
long long Long() {
  long long x;
  scanf("%I64d", &x);
  return x;
}
const int N = (int)1e5 + 2;
const long long MOD = (int)1e9 + 9;
int main() {
  int n = Int(), k = Int();
  if (k > n) return puts("NO"), 0;
  priority_queue<int> st;
  while (n) {
    int p = log2(n);
    int m = (int)powl(2LL, p * 1LL);
    st.push(m);
    n -= m;
  }
  int mn = (int)st.size();
  if (k < mn) return puts("NO"), 0;
  puts("YES");
  while ((int)st.size() < k) {
    int nw = st.top();
    st.pop();
    st.push(nw / 2);
    st.push(nw / 2);
  }
  while (st.size()) {
    cout << st.top() << " ";
    st.pop();
  }
  cout << '\n';
  return 0;
}