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;
long long BigMod(long long int a, long long int b, int M) {
  long long remainder, answer = 1;
  remainder = a % M;
  while (b != 0) {
    if (b % 2 == 1) answer = (answer * remainder) % M;
    remainder = (remainder * remainder) % M;
    b /= 2;
  }
  return answer;
}
bool cmp(int a, int b) { return a > b; }
int main() {
  int n, k;
  cin >> n >> k;
  vector<int> v;
  while (n) {
    v.push_back(n & 1);
    if (n & 1) k--;
    n >>= 1;
  }
  if (k < 0) {
    printf("NO\n");
    return 0;
  }
  int tmp;
  for (int i = v.size() - 1; i > 0; i--) {
    tmp = min(v[i], k);
    v[i] -= tmp;
    v[i - 1] += 2 * tmp;
    k -= tmp;
  }
  if (k > 0) {
    printf("NO\n");
    return 0;
  }
  printf("YES\n");
  for (int i = 0, cnt = 1; i < v.size(); i++, cnt <<= 1) {
    for (int j = 0; j < v[i]; j++) {
      printf("%d ", cnt);
    }
  }
  return 0;
}