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;
void printv(vector<long long> v) {
  for (long long i = 0; i < (long long)v.size(); i++) {
    cout << v[i] << ' ';
  }
  cout << '\n';
}
long long be(long long m, long long n) {
  long long res = 1;
  while (n) {
    if (n % 2 != 0) {
      res *= m;
    }
    m *= m;
    n /= 2;
  }
  return res;
}
void sieve(long long n, vector<bool> &prime) {
  prime[0] = false;
  prime[1] = false;
  for (long long i = 2; i * i <= n; i++) {
    if (prime[i] == true) {
      for (long long j = i * i; j <= n; j += i) prime[j] = false;
    }
  }
}
void solve() {
  long long n, k;
  cin >> n >> k;
  priority_queue<long long> pq;
  long long i = 0;
  while (n) {
    if (n & 1) {
      pq.push(be(2, i));
    }
    n /= 2;
    i++;
  }
  if (pq.size() > k) {
    cout << "NO\n";
    return;
  }
  while (pq.size() < k) {
    long long z = pq.top();
    pq.pop();
    if (z == 1) break;
    pq.push(z / 2);
    pq.push(z / 2);
  }
  if (pq.size() < k) {
    cout << "NO\n";
    return;
  }
  cout << "YES\n";
  while (!pq.empty()) {
    cout << pq.top() << ' ';
    pq.pop();
  }
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long t = 1;
  while (t--) {
    solve();
  }
}