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;
const int N = 2e5 + 5, OO = 0x3f3f3f3f;
int n, k, pow2[N];
void _run() {
  scanf("%d %d", &n, &k);
  pow2[0] = 1;
  for (int i = 1; i <= 31; ++i) pow2[i] = pow2[i - 1] * 2;
  int idx = upper_bound(pow2, pow2 + 32, n) - pow2;
  --idx;
  while (idx >= 0) {
    int tmpRkm = n, tmpK = k, tmpIdx = idx;
    vector<int> ans;
    while (tmpRkm != 0 && tmpK && tmpIdx > -1) {
      if (tmpRkm - pow2[tmpIdx] >= 0 && tmpRkm - pow2[tmpIdx] >= tmpK - 1) {
        ans.push_back(pow2[tmpIdx]);
        tmpRkm -= pow2[tmpIdx];
        tmpK--;
      } else
        tmpIdx--;
    }
    if (tmpRkm == 0 && tmpK == 0) {
      puts("YES");
      for (auto x : ans) printf("%d ", x);
      return;
    }
    idx--;
  }
  puts("NO");
}
int main() {
  int tc = 1;
  while (tc--) _run();
  return 0;
}