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> binR;
int findIndex() {
  for (int x = 0; x < binR.size(); x++) {
    if (binR.at(x) > 0) {
      return x;
    }
  }
  return -1;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int n, k;
  cin >> n >> k;
  int temp = n;
  int val = 0;
  if (temp == 0) {
    binR.push_back(0);
  }
  while (temp >= pow(2, val)) {
    val++;
  }
  for (int x = val - 1; x >= 0; --x) {
    int curr = pow(2, x);
    if (temp >= curr) {
      temp -= curr;
      binR.push_back(1);
    } else {
      binR.push_back(0);
    }
  }
  int i = 0;
  for (auto x : binR) {
    if (x == 1) i++;
  }
  if (k < i || k > n) {
    cout << "NO";
  } else {
    while (i < k) {
      int loc = findIndex();
      binR.at(loc)--;
      binR.at(loc + 1) += 2;
      i++;
    }
    cout << "YES\n";
    cout << setprecision(12);
    while (true) {
      int loc = findIndex();
      if (loc == -1) break;
      cout << pow(2, binR.size() - loc - 1) << " ";
      binR.at(loc)--;
    }
  }
  return 0;
}