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 long long mod = 1e9 + 7;
const long long N = 1e5 + 5;
string s, s1, s2;
vector<long long> v, v1, v2;
vector<int> g[N];
map<long long, long long> mp;
long long n, m, cnt, k, y, x, t, ans, mn = mod, mx = mod, z;
int used[N];
int main() {
  ios_base::sync_with_stdio(0);
  long long a, b, c;
  cin >> a >> b;
  x = a;
  for (int i = 1; i <= a; i *= 2) {
    v.push_back(i);
  }
  if (a != v[v.size() - 1] and b == 1) {
    cout << "NO" << endl;
    ;
    return 0;
    ;
  }
  reverse(v.begin(), v.end());
  int i = 0;
  while (i < v.size()) {
    if (a < v[i]) i++;
    cnt = a - v[i];
    if (cnt > b - 1 and v[i] != a) {
      v1.push_back(v[i]);
      a -= v[i];
      b--;
    } else if (cnt == b - 1) {
      v1.insert(v1.begin(), b - 1, 1);
      v1.push_back(v[i]);
      break;
    } else
      i++;
  }
  cnt = 0;
  reverse(v1.begin(), v1.end());
  for (auto it : v1) cnt += it;
  if (cnt != x) return cout << "NO", 0;
  cout << "YES\n";
  for (auto it : v1) cout << it << ' ';
}