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 double pi = acos(-1.0);
const double eps = 1e-11;
template <class T>
inline void checkmin(T& a, T b) {
  if (b < a) a = b;
}
template <class T>
inline void checkmax(T& a, T b) {
  if (b > a) a = b;
}
template <class T>
inline T sqr(T x) {
  return x * x;
}
template <class T>
inline T lsb(T n) {
  return (n & (-n));
}
template <class T>
inline int countbit(T n) {
  return (n == 0) ? 0 : (1 + countbit(n & (n - 1)));
}
template <class T>
inline T gcd(T a, T b) {
  if (a < 0) return gcd(-a, b);
  if (b < 0) return gcd(a, -b);
  return (b == 0) ? a : gcd(b, a % b);
}
bool isequal(vector<int> a, vector<int> b) {
  for (int i = 0; i < (a.size()); i++)
    if (a[i] != b[i]) return false;
  return true;
}
template <class T>
bool allequal(vector<T> a) {
  sort(a.begin(), a.end());
  return a[0] == a[a.size() - 1];
}
template <class T>
void printarray(vector<T> a) {
  for (int i = 0; i < (a.size()); i++) cout << a[i] << " ";
  cout << endl;
}
template <class T>
void printsubarray(vector<T> a, int start, int end) {
  if (start < 0 or end > a.size() - 1) cout << "Out of bounds" << endl;
  for (int i = start; i <= end; i++) cout << a[i] << " ";
  cout << endl;
}
bool isprime(int n) {
  if (n < 2) return false;
  if (n == 2) return true;
  for (int i = 2; i * i <= n; i++) {
    if (n % i == 0) return false;
  }
  return true;
}
void split(const string& s, char c, vector<string>& result) {
  stringstream ss(s);
  string x;
  while (getline(ss, x, c)) result.push_back(x);
}
void err(vector<string>::iterator it) {}
template <typename T, typename... Args>
void err(vector<string>::iterator it, T a, Args... args) {
  cerr << it->substr((*it)[0] == ' ', it->length()) << " = " << a << '\n';
  err(++it, args...);
}
void readarray(vector<int>& a, int size) {
  int tmp;
  a.clear();
  for (int i = 0; i < (size); i++) {
    cin >> tmp;
    a.push_back(tmp);
  }
}
bool invalid(int n, int k) {
  int nbits = countbit(n);
  if (k < nbits or k > n) return true;
  return false;
}
vector<int> res;
void split(int n, int k) {
  if (k == 1) {
    res.push_back(n);
  } else {
    int nbits = countbit(n);
    if (nbits == 1) {
      split(n >> 1, ((k >> 1) + k % 2));
      split(n >> 1, (k >> 1));
    } else {
      int a, b;
      int msb = (int)log2(n);
      a = (1 << msb);
      b = n ^ a;
      int excess = k - nbits + 1;
      int first = min(excess, a);
      {
        vector<string> _v;
        split("a, b, excess, nbits, k, first", ',', _v);
        err(_v.begin(), a, b, excess, nbits, k, first);
      };
      split(a, first);
      split(b, k - first);
    }
  }
}
void solvecase() {
  int n, k;
  cin >> n >> k;
  if (invalid(n, k)) {
    cout << "NO" << endl;
    return;
  }
  split(n, k);
  cout << "YES" << endl;
  printarray(res);
}
int main() {
  int cases = 1;
  while (cases--) {
    solvecase();
  }
  return 0;
}