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 gcd(long long a, long long b) {
  if (a == 0) return b;
  return gcd(b % a, a);
}
bool mod(long double a, long double b) { return a / b - floor(a / b); }
long long occurs(vector<long long> v, long long n) {
  auto it = lower_bound(((v)).begin(), ((v)).end(), (n));
  auto it1 = upper_bound(((v)).begin(), ((v)).end(), (n));
  long long first = it1 - it;
  return first;
}
long long logb(long long base, long long first) {
  return (log(first) / log(base));
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long i, j, first, n, m, c = 0, d, a, b;
  long long arr[35];
  first = 0;
  cin >> n >> d;
  m = n;
  while (n > 0) {
    if (n % 2 == 1) c++;
    arr[first] = n % 2;
    n /= 2;
    first++;
  }
  n = m;
  first--;
  if (d < c)
    cout << "NO";
  else if (d > n)
    cout << "NO";
  else {
    for (i = first; i >= 1; i--) {
      if (c == d) break;
      a = arr[i];
      if ((c + a) <= d) {
        arr[i] = 0;
        c += a;
        arr[i - 1] += (2 * a);
      } else {
        b = d - c;
        arr[i] -= b;
        arr[i - 1] += 2 * b;
        break;
      }
    }
    cout << "YES" << '\n';
    for (i = 0; i < first + 1; i++) {
      a = 1 << i;
      b = arr[i];
      for (j = 0; j < arr[i]; j++) {
        cout << a << " ";
      }
    }
  }
  return 0;
}