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>
#pragma GCC optimize("unroll-loops", "omit-frame-pointer", "inline")
#pragma GCC option("arch=native", "tune=native", "no-zero-upper")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,avx2")
#pragma GCC optimize("Ofast")
using namespace std;
const long double pi = acos(-1);
const long long int M = 1e9 + 7, N = 1e7 + 5, INF = 1e18;
void solve() {
  long long int n, k, nn, d;
  cin >> n >> k;
  vector<long long int> bn;
  nn = n;
  while (n > 0) {
    bn.push_back(n % 2);
    n /= 2;
  }
  reverse(bn.begin(), bn.end());
  d = k - __builtin_popcount(nn);
  if (k < __builtin_popcount(nn)) {
    cout << "NO"
         << "\n";
  } else {
    for (long long int i = 0; i <= (long long int)(bn).size() - 2; i++) {
      while (bn[i] && d) {
        bn[i]--;
        bn[i + 1] += 2;
        d--;
      }
    }
    long long int p = 1;
    if (d != 0) {
      cout << "NO"
           << "\n";
    } else {
      cout << "YES"
           << "\n";
      for (long long int i = (long long int)(bn).size() - 1; i >= 0; i--) {
        while (bn[i]--) {
          cout << p << " ";
        }
        p *= 2;
      }
    }
  }
}
int32_t main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  solve();
  return 0;
}