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;
int block(long int x) {
  vector<long int> v;
  while (x > 0) {
    v.push_back(x % 2);
    x = x / 2;
  }
  int ret = 0;
  for (int i = 0; i < v.size(); i++) {
    if (v[i] == 1) {
      ret++;
    }
  }
  return ret;
}
void pprint(long int x) {
  vector<long int> v;
  while (x > 0) {
    v.push_back(x % 2);
    x = x / 2;
  }
  cout << "YES\n";
  int ret = 0;
  for (int i = 0; i < v.size(); i++) {
    if (v[i] == 1) {
      ret++;
      int A = 1 << i;
      cout << A << " ";
    }
  }
  cout << endl;
}
void solve(long int x, long long k, long long mini) {
  vector<long int> v;
  long long rem = k - mini;
  while (x > 0) {
    v.push_back(x % 2);
    x = x / 2;
  }
  int ret = 0;
  cout << "YES\n";
  int i = v.size() - 1;
  while (rem > 0) {
    v[i]--;
    v[i - 1] += 2;
    if (v[i] == 0) {
      i--;
    }
    rem--;
  }
  for (int i = 0; i < v.size(); i++) {
    while (v[i] > 0) {
      ret++;
      int A = 1 << i;
      cout << A << " ";
      v[i]--;
    }
  }
  cout << endl;
}
int main() {
  long long x;
  long long n, k;
  cin >> n >> k;
  if (k > n) {
    cout << "NO\n";
    return 0;
  }
  if (k == n) {
    cout << "YES\n";
    for (int i = 0; i < n; i++) {
      cout << 1 << " ";
    }
    return 0;
  }
  int mini = block(n);
  if (k < mini) {
    cout << "NO\n";
    return 0;
  }
  if (k == mini) {
    pprint(n);
  } else {
    solve(n, k, mini);
  }
  return 0;
}