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 n, k, i, temp = 0;
int a[100];
int main() {
  cin >> n >> k;
  int t = n;
  while (t != 0) {
    a[i++] = t % 2;
    t >>= 1;
    if (a[i - 1] == 1) temp++;
  }
  if (temp == k) {
    cout << "YES" << endl;
    for (int j = 0; j < i; j++)
      if (a[j] == 1) {
        cout << (int)pow(2, j) << "  ";
      }
    return 0;
  }
  if (temp > k || n < k) {
    cout << "NO" << endl;
    return 0;
  }
  int num = i - 1;
  while (temp != k) {
    a[num]--;
    a[num - 1] += 2;
    if (a[num] == 0) num--;
    temp++;
  }
  cout << "YES" << endl;
  for (int j = 0; j < i; j++)
    while (a[j] != 0) {
      cout << (int)pow(2, j) << "  ";
      a[j]--;
    }
  return 0;
}