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;
long long int exp(long long int a, long long int b, long long int m);
long long int po(long long int n) {
  long long int co = 0;
  while (n / 2 > 0) {
    n /= 2;
    co++;
  }
  return co;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long int n, k;
  cin >> n >> k;
  long long int kk = k;
  long long int nn = n;
  vector<long long int> pr;
  long long int k1 = 0;
  bool ans = 1;
  while (n > 0 && k > 0) {
    long long int pp = po(n);
    long long int pt = exp(2, pp, 1000000007);
    while (n - pt < k - 1 && pt > 0) {
      pt /= 2;
    }
    if (pt == 0) {
      ans == 0;
      break;
    }
    k--;
    n -= pt;
    pr.push_back(pt);
  }
  if (n == 0 && pr.size() == kk) {
    long long int sum = 0;
    for (long long int i = 0; i < kk; i++) {
      sum += pr[i];
    }
    if (sum == nn) {
      cout << "YES" << endl;
      for (long long int i = 0; i < kk; i++) cout << pr[i] << " ";
      cout << endl;
    } else {
      cout << "NO" << endl;
    }
  } else {
    cout << "NO" << endl;
  }
  return 0;
}
long long int exp(long long int a, long long int b, long long int m) {
  if (b == 0) {
    return 1;
  }
  long long int temp = exp(a, b / 2, m);
  temp = (temp * temp) % m;
  if (b & 1) {
    return (temp * (a % m)) % m;
  }
  return temp;
}