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;
int index(int numToFind, std::vector<int> vectorToFindIn) {
  int len = vectorToFindIn.size();
  for (int i = 0; i < len; i++) {
    if (vectorToFindIn[i] == numToFind) {
      return i;
    }
  }
  return -1;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int n, k, i, power, j;
  power = 536870912;
  cin >> n >> k;
  int a[k];
  i = 0;
  if (k > n) {
    cout << "NO";
    return 0;
  }
  if (n == 2 * 536870912) {
    a[0] = 2 * 536870912;
    ++i;
    n = 0;
  }
  while (n > 0 && i < k) {
    if (n >= power) {
      n -= power;
      a[i] = power;
      ++i;
    }
    power /= 2;
  }
  if (i <= k && n == 0) {
    j = 0;
    while (i < k) {
      while (i < k && a[j] > 1) {
        a[i] = a[j] / 2;
        a[j] /= 2;
        ++i;
      }
      ++j;
    }
    cout << "YES\n";
    for (int m = 0; m < k; m++) {
      cout << a[m] << " ";
    }
  } else {
    cout << "NO";
  }
  return 0;
}