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;
const long long int MOD = 998244353;
const int N = 200005;
const long long int p = 53;
int dx[] = {-1, -1, -1, 0, 0, 1, 1, 1};
int dy[] = {-1, 0, 1, -1, 1, -1, 0, 1};
void testCase() {
  long long int n, k;
  cin >> n >> k;
  bitset<32> bt(n);
  long long int kk = k;
  long long int mx = bt.count();
  if (k < mx) {
    cout << "NO";
    return;
  }
  multiset<long long int> mst;
  for (long long int i = 0; i < 32; i++) {
    if (bt[i] == 1) {
      mst.insert(pow(2, i));
      k--;
    }
  }
  while (k > 0) {
    long long int x = *mst.rbegin();
    if (x == 1) {
      break;
    }
    mst.insert(x / 2);
    mst.insert(x / 2);
    auto itr = mst.end();
    --itr;
    mst.erase(itr);
    k--;
  }
  if (mst.size() < kk) {
    cout << "NO";
    return;
  }
  cout << "YES\n";
  for (auto itr : mst) {
    cout << itr << " ";
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int t = 1;
  while (t != 0) {
    t--;
    testCase();
  }
  return 0;
}