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;
long long power(long long x, long long y, long long p) {
  long long res = 1;
  x = x % p;
  while (y > 0) {
    if (y & 1) res = (res * x) % p;
    y = y >> 1;
    x = (x * x) % p;
  }
  return res;
}
long long ceil(long long a, long long b) {
  if (a % b == 0) return a / b;
  return a / b + 1;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long n, k;
  cin >> n >> k;
  long long temp = n;
  vector<long long> ans;
  long long count = 0;
  if (k > temp) {
    cout << "NO" << endl;
    return 0;
  }
  while (k < ceil(temp, 2) && temp > 0) {
    if (temp % 2 == 1) {
      k--;
      ans.push_back(pow(2, count));
    }
    count++;
    temp /= 2;
  }
  long long i;
  vector<long long> finalans;
  for (i = 0; i < ans.size(); i++) finalans.push_back(ans[i]);
  for (i = 1; i <= (temp - k); i++) finalans.push_back(pow(2, count + 1));
  for (i = 1; i <= (2 * k - temp); i++) finalans.push_back(pow(2, count));
  long long check = 0;
  for (i = 0; i < finalans.size(); i++) check += finalans[i];
  if (check != n)
    cout << "NO" << endl;
  else {
    cout << "YES" << endl;
    for (i = 0; i < finalans.size(); i++) cout << finalans[i] << ' ';
    cout << endl;
  }
  return 0;
}