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;
using ll = long long;
using db = double;
using ld = long double;
template <typename T>
using V = vector<T>;
template <typename T>
using VV = vector<vector<T>>;
const ll mod = 1000000007;
const ll INF = 1000000099;
const ll LINF = (ll)(1e18 + 99);
vector<ll> dx = {-1, 1, 0, 0}, dy = {0, 0, -1, 1};
template <typename T, typename U>
void chmin(T& t, const U& u) {
  if (t > u) t = u;
}
template <typename T, typename U>
void chmax(T& t, const U& u) {
  if (t < u) t = u;
}
template <typename T>
T gcd(T a, T b) {
  return b ? gcd(b, a % b) : a;
}
template <typename T>
T mpow(T a, T n) {
  T res = 1;
  for (; n; n >>= 1) {
    if (n & 1) res = res * a;
    a = a * a;
  }
  return res;
}
signed main() {
  cin.tie(0);
  ios::sync_with_stdio(false);
  cout << fixed << setprecision(20);
  ll n, k;
  cin >> n >> k;
  if (__builtin_popcount(n) > k || n < k) {
    cout << "NO" << endl;
    return 0;
  } else {
    cout << "YES" << endl;
    k -= __builtin_popcount(n);
  }
  ll exp = 0;
  queue<ll> q;
  while (n > 0) {
    if (n % 2) q.push(mpow(2ll, exp));
    exp++;
    n >>= 1;
  }
  while (q.size()) {
    ll now = q.front();
    q.pop();
    if (k > 0 && now > 1) {
      q.push(now / 2);
      q.push(now / 2);
      k--;
    } else {
      cout << now << " ";
    }
  }
}