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 n, k, cnt = 0;
list<int> powersOf2;
map<int, int> appearance;
void fillBits(int a) {
  int pos = -1;
  while (a) {
    pos++;
    if ((a & 1)) {
      appearance[1 << pos]++;
      if (pos) powersOf2.push_back(1 << pos);
    }
    a >>= 1;
  }
}
signed main() {
  scanf("%d %d", &n, &k);
  fillBits(n);
  cnt = appearance.size();
  if (cnt > k) {
    printf("NO\n");
    return 0;
  }
  while (cnt < k && !powersOf2.empty()) {
    int back = powersOf2.back();
    powersOf2.pop_back();
    appearance[back / 2] += 2;
    appearance[back]--;
    if (back / 2 > 1) {
      powersOf2.push_back(back / 2);
      powersOf2.push_back(back / 2);
    }
    cnt++;
  }
  if (cnt < k) {
    printf("NO\n");
    return 0;
  }
  printf("YES\n");
  for (const auto &it : appearance) {
    for (int j = 0; j < it.second; j++) {
      printf("%d ", it.first);
    }
  }
  printf("\n");
}