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;
vector<int> b;
int s;
void func(long long int n) {
  while (n > 0) {
    int j = n % 2;
    if (j == 1) s++;
    b.push_back(j);
    n = n / 2;
  }
}
int main() {
  long long int n, k, i, j, l;
  s = 0;
  cin >> n >> k;
  func(n);
  if (s > k) {
    cout << "NO" << endl;
  } else if (s == k) {
    long long int f = 1;
    cout << "YES" << endl;
    for (i = 0; i < b.size(); i++) {
      for (j = 0; j < b[i]; j++) {
        cout << f << " ";
      }
      f = f * 2;
    }
  } else {
    for (i = b.size() - 1; i > 0 && s < k; i--) {
      if (b[i] + s <= k) {
        b[i - 1] = b[i - 1] + 2 * b[i];
        s = s + b[i];
        b[i] = 0;
      } else {
        l = k - s;
        b[i - 1] = b[i - 1] + 2 * l;
        s = s + l;
        b[i] = b[i] - l;
      }
    }
    if (s != k) {
      cout << "NO" << endl;
      return 0;
    }
    cout << "YES" << endl;
    long long int f = 1;
    for (i = 0; i < b.size(); i++) {
      for (j = 0; j < b[i]; j++) {
        cout << f << " ";
      }
      f = f * 2;
    }
  }
  return 0;
}