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 main(void) {
  int n, k, v, i = 0, c = 0;
  cin >> n >> k;
  int arr[50] = {0};
  if (n < k) {
    cout << "NO" << endl;
    return 0;
  }
  if (n == k) {
    cout << "YES" << endl;
    int t = 1;
    while (t <= n) {
      cout << "1 ";
      t++;
    }
    cout << endl;
    return 0;
  } else {
    int a = n, rem;
    while (a > 0) {
      rem = a % 2;
      if (rem == 1) {
        ++c;
      }
      ++i;
      arr[i] = rem;
      a = a / 2;
    }
  }
  v = i;
  if (c > k) {
    cout << "NO" << endl;
    return 0;
  } else {
    int count = c;
    while (count < k) {
      while (arr[i] == 0) {
        --i;
      }
      --arr[i];
      arr[i - 1] += 2;
      ++count;
    }
  }
  cout << "YES" << endl;
  int p;
  for (int j = 1; j <= v; j++) {
    if (arr[j] != 0) {
      p = pow(2, j - 1);
      int t = 1;
      while (t <= arr[j]) {
        cout << p << " ";
        ++t;
      }
    }
  }
  cout << endl;
  return 0;
}