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;
bool isPowerOfTwo(long long n) { return (ceil(log2(n)) == floor(log2(n))); }
long long binpow(long long a, long long b) {
  long long res = 1;
  while (b > 0) {
    if (b & 1) res *= a;
    a = a * a;
    b >>= 1;
  }
  return res;
}
bool sortbysec(const pair<int, int> &a, const pair<int, int> &b) {
  bool x = false;
  if ((a.second * a.first) < b.second * b.first) {
    x = true;
  }
  return x;
}
int main() {
  long long n, k;
  cin >> n >> k;
  long long c = 1;
  long long nn = n;
  long long t = 0;
  vector<long long> v;
  while (n > 0) {
    if (n & 1) {
      t++;
      v.push_back(c);
    }
    c *= 2;
    n >>= 1;
  }
  if (k >= t && k <= nn) {
    cout << "YES"
         << "\n";
    if (k == t) {
      for (auto x : v) cout << x << " ";
      cout << "\n";
      return 0;
    }
    if (k == nn) {
      for (int i = 0; i < nn; i++) cout << 1 << " ";
      cout << "\n";
      return 0;
    }
    k = k - v.size();
    while (k > 0) {
      for (int i = 0; i < v.size(); i++) {
        if (k == 0) {
          break;
        }
        if (v[i] > 1) {
          v.push_back(v[i] / 2);
          v[i] = (v[i] / 2);
          k--;
        }
      }
    }
    for (auto x : v) cout << x << " ";
    cout << "\n";
    return 0;
  } else
    cout << "NO"
         << "\n";
}