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;
inline int min(int a, int b) { return a > b ? b : a; }
inline int max(int a, int b) { return a > b ? a : b; }
inline int abs(int a) { return a >= 0 ? a : -a; }
int maxArr(vector<int>& a) {
  if (a.size() == 0) return -1;
  int ind = 0;
  for (long long i = 0; i < a.size(); i++) {
    if (a[i] > a[ind]) {
      ind = i;
    }
  }
  return a[ind];
}
int minArr(vector<int>& a) {
  if (a.size() == 0) return -1;
  int ind = 0;
  for (long long i = 0; i < a.size(); i++) {
    if (a[i] < a[ind]) {
      ind = i;
    }
  }
  return a[ind];
}
long long sumArr(vector<int>& a) {
  long long sum = 0;
  for (long long i = 0; i < a.size(); i++) {
    sum += a[i];
  }
  return sum;
}
int main() {
  int n, k;
  cin >> n >> k;
  vector<int> a = vector<int>();
  if (k > n) {
    cout << "NO\n";
    return 0;
  }
  while (n != 0) {
    a.push_back(n % 2);
    n /= 2;
  }
  if (k < sumArr(a)) {
    cout << "NO\n";
    return 0;
  }
  int s = sumArr(a);
  for (int i = a.size() - 1; i >= 1; i--) {
    if (s == k) break;
    if (s + a[i] > k) {
      a[i - 1] += 2 * (k - s);
      a[i] -= (k - s);
      s = k;
    } else {
      s += a[i];
      a[i - 1] += a[i] * 2;
      a[i] = 0;
    }
  }
  cout << "YES" << endl;
  for (long long i = 0; i < a.size(); i++) {
    for (long long j = 0; j < a[i]; j++) {
      cout << (1 << i) << " ";
    }
  }
}