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 int mxN = 1e6 + 15;
const int mod = 1e9 + 7;
const long long INF = 1e9;
using Matrix = vector<vector<long long>>;
using vll = vector<long long>;
using cd = complex<double>;
const double PI = acos(-1);
clock_t begtime = clock();
int bit[100];
void solve() {
  long long n, k;
  cin >> n >> k;
  string s;
  long long mn = 0, mx = 0;
  map<int, int> mp;
  queue<int> q;
  for (int i = 0; i <= 30; i++) {
    if (n & (1 << i)) {
      mx += (1 << i);
      mp[(1 << i)]++;
      if (i > 0) q.push((1 << i));
      mn++;
    }
  }
  if (k >= mn && k <= mx) {
    cout << "YES" << endl;
  } else {
    cout << "NO" << endl;
    return;
  }
  long long cnt = mn;
  while (cnt < k) {
    int z = q.front();
    q.pop();
    mp[z]--;
    mp[z / 2] += 2;
    if (z / 2 > 1) {
      q.push(z / 2);
      q.push(z / 2);
    }
    cnt++;
  }
  for (auto x : mp) {
    for (int i = 0; i < x.second; i++) {
      cout << x.first << " ";
    }
  }
}
signed main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout << setprecision(10) << fixed;
  long long t = 1;
  while (t--) {
    solve();
  }
  return 0;
}