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 mod1 = 1e17 + 3;
long long mod = 998244353;
const long long N = 1e5 + 10;
long long n, k;
long long m[50];
int func(int x) {
  int i = 0, j = 0;
  while (x) {
    if (x % 2) i++, m[j]++;
    j++;
    x /= 2;
  }
  return i;
}
void solve() {
  cin >> n >> k;
  if (k > n) {
    cout << "NO";
    return;
  }
  int x = func(n);
  if (x > k) {
    cout << "NO";
    return;
  }
  k -= x;
  for (int i = 49; i >= 0; i--) {
    if (m[i] == 0 || k == 0) continue;
    if (m[i] <= k)
      k -= m[i], m[i - 1] += 2 * m[i], m[i] = 0;
    else {
      m[i] -= k;
      m[i - 1] += 2 * k;
      k = 0;
    }
  }
  cout << "YES"
       << "\n";
  for (int i = 0; i < 40; i++) {
    while (m[i]--) {
      cout << setprecision(30) << pow(2, i) << " ";
    }
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  ;
  int t = 1;
  while (t--) {
    solve();
    cout << "\n";
  }
  return 0;
}