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 maxn = 2e5 + 10;
int n, k, t, p = 1;
int ans[maxn], top, pointer;
int main() {
  while (cin >> n >> k) {
    pointer = t = top = 0;
    p = 1;
    memset(ans, 0, sizeof(ans));
    for (int i = n; i; i >>= 1, p <<= 1)
      if (i & 1) t++, ans[++top] = p;
    if (t > k || n < k) {
      cout << "NO" << endl;
      continue;
    }
    cout << "YES" << endl;
    t = k - t;
    pointer = top;
    while (t > 0) {
      if (ans[pointer] != 1) ans[pointer] >>= 1, ans[++top] = ans[pointer];
      while (ans[pointer] == 1 && pointer > 0) pointer--;
      if (pointer == 0) pointer = top;
      while (ans[pointer] == 1 && pointer > 0) pointer--;
      t--;
    }
    sort(ans + 1, ans + top + 1);
    for (int i = 1; i <= top; i++) cout << ans[i] << " ";
    cout << endl;
  }
  return 0;
}