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 namespace chrono;
void __print(long long x) { cerr << x; }
void __print(float x) { cerr << x; }
void __print(double x) { cerr << x; }
void __print(long double x) { cerr << x; }
void __print(char x) { cerr << '\'' << x << '\''; }
void __print(const char *x) { cerr << '\"' << x << '\"'; }
void __print(const string &x) { cerr << '\"' << x << '\"'; }
void __print(bool x) { cerr << (x ? "true" : "false"); }
template <typename T, typename V>
void __print(const pair<T, V> &x) {
  cerr << '{';
  __print(x.first);
  cerr << ',';
  __print(x.second);
  cerr << '}';
}
template <typename T>
void __print(const T &x) {
  long long f = 0;
  cerr << '{';
  for (auto &i : x) cerr << (f++ ? "," : ""), __print(i);
  cerr << "}";
}
void _print() { cerr << "]\n"; }
template <typename T, typename... V>
void _print(T t, V... v) {
  __print(t);
  if (sizeof...(v)) cerr << ", ";
  _print(v...);
}
void inout() {}
template <typename T>
void showvector(vector<T> vec) {
  for (auto elem : vec) cout << elem << " ";
  cout << "\n";
}
template <typename T>
void inputvector(vector<T> &vec) {
  for (auto &x : vec) cin >> x;
}
void precomputation() {}
bool ispossible(long long n, long long k) {
  long long cnt = 0;
  for (long long i = 0; i < 31; i++) {
    if ((n & (1 << i)) >= 1) cnt++;
  }
  if (k < cnt || k > n)
    return 0;
  else
    return 1;
}
long long power(long long x, long long y) {
  if (y == 0) return 1;
  if (y % 2 == 0)
    return power(x, y / 2) * power(x, y / 2);
  else
    return x * power(x, (y - 1) / 2) * power(x, (y - 1) / 2);
}
void solve() {
  long long n, k;
  vector<long long> v(31), v1;
  for (long long i = 0; i < 31; i++) {
    v[i] = power(2, i);
  }
  cin >> n >> k;
  bool ans = ispossible(n, k);
  ;
  if (ans == 0) {
    cout << "NO"
         << "\n";
    return;
  };
  long long n1 = n;
  while (n1) {
    long long index = -1;
    for (long long i = 0; i < 31; i++) {
      if (v[i] > n1) {
        index = i - 1;
        break;
      }
    }
    for (long long i = index; i >= 0; i--) {
      if (ispossible(n1 - v[i], k - 1)) {
        v1.push_back(v[i]);
        k--;
        n1 -= v[i];
        break;
      }
    }
  }
  cout << "YES"
       << "\n";
  for (auto x : v1) {
    cout << x << " ";
  }
  cout << "\n";
}
signed main() {
  inout();
  ios::sync_with_stdio(0);
  cin.tie(0);
  precomputation();
  { solve(); }
  return 0;
}