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;
double eps = 0.0000001;
long long fastpower(long long b, long long p) {
  long long ans = 1;
  while (p) {
    if (p % 2) {
      ans = (ans * b);
    }
    b = b * b;
    p /= 2;
  }
  return ans;
}
long long fastpowermod(long long b, long long p, long long m) {
  long long ans = 1;
  while (p) {
    if (p % 2) {
      ans = ((ans % m) * (b % m)) % m;
    }
    b = ((b % m) * (b % m)) % m;
    p /= 2;
  }
  return ans % m;
}
string makeitstring(long long n) {
  string ans;
  bool ch = 0;
  if (n < 0) {
    n *= -1;
    ch = 1;
  }
  if (n == 0) {
    ans = "0";
  };
  while (n) {
    long long mo = n % 10;
    mo += 48;
    char m = mo;
    ans = m + ans;
    n /= 10;
  }
  if (ch) {
    ans = '-' + ans;
  }
  return ans;
}
bool compare(double f, double s) { return (abs(f - s) < eps); }
long long gcd(long long a, long long b) {
  while (b != 0) {
    long long a2 = a;
    a = b;
    b = a2 % b;
  }
  return a;
}
long long wanted;
void solve(long long n) {
  priority_queue<int> p;
  int now = 0;
  int cnt = 0;
  while (n) {
    if (n % 2) {
      p.push(fastpower(2, cnt));
    }
    n /= 2;
    cnt++;
  }
  while (p.size() < wanted) {
    int cur = p.top();
    if (cur == 1) {
      break;
    }
    cur /= 2;
    p.push(cur);
    p.push(cur);
    p.pop();
  }
  if (p.size() != wanted) {
    cout << "NO";
    return;
  }
  cout << "YES" << endl;
  while (p.size()) {
    cout << p.top() << " ";
    p.pop();
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  ;
  long long n;
  cin >> n >> wanted;
  solve(n);
  return 0;
}