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 a[50][50];
bool feel(long long x, long long y) {
  if (a[x][y] == 0) return false;
  if (a[x][y + 1] == 0) return false;
  if (a[x + 1][y] == 0) return false;
  if (a[x + 1][y + 1] == 0) return false;
  a[x][y] = -1;
  a[x + 1][y] = -1;
  a[x][y + 1] = -1;
  a[x + 1][y + 1] = -1;
  return true;
}
void solve() {
  long long n, k;
  cin >> n >> k;
  long long num = __builtin_popcountll(n);
  if (k < num || k > n) {
    cout << "NO";
    return;
  }
  multiset<long long> v;
  long long j = 1;
  while (n > 0) {
    if (n % 2) v.insert(j);
    n = n / 2;
    j = j * 2;
  }
  while (num < k) {
    auto itr = v.end();
    itr--;
    long long temp = *itr;
    temp = temp / 2;
    v.erase(itr);
    v.insert(temp);
    v.insert(temp);
    num++;
  }
  cout << "YES" << '\n';
  auto itr = v.begin();
  while (itr != v.end()) {
    cout << *itr << " ";
    itr++;
  }
  return;
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  ;
  long long n = 1;
  for (long long i = 0; i < n; i++) solve();
  return 0;
}