description: You are given an array a consisting of n distinct elements and an integer k. Each element in the array is a non-negative integer not exceeding 2^k-1.
  
  Let's define the XOR distance for a number x as the value of 
  
  $$$f(x) = min_{i = 1}^{n} min_{j = i + 1}^{n} |(a_i ⊕ x) - (a_j ⊕ x)|,$$$
  
  where ⊕ denotes [the bitwise XOR operation](https://en.wikipedia.org/wiki/Bitwise_operation#XOR).
  
  For every integer x from 0 to 2^k-1, you have to calculate f(x).
  
  Input
  
  The first line contains two integers n and k (1 ≤ k ≤ 19; 2 ≤ n ≤ 2^k).
  
  The second line contains n integers a_1, a_2, ..., a_n (0 ≤ a_i ≤ 2^k-1). All these integers are distinct.
  
  Output
  
  Print 2^k integers. The i-th of them should be equal to f(i-1).
  
  Examples
  
  Input
  
  
  3 3
  6 0 3
  
  
  Output
  
  
  3 1 1 2 2 1 1 3 
  
  
  Input
  
  
  3 4
  13 4 2
  
  
  Output
  
  
  2 2 6 6 3 1 2 2 2 2 1 3 6 6 2 2 
  
  Note
  
  Consider the first example:
  
    * for x = 0, if we apply bitwise XOR to the elements of the array with x, we get the array [6, 0, 3], and the minimum absolute difference of two elements is 3; 
    * for x = 1, if we apply bitwise XOR to the elements of the array with x, we get the array [7, 1, 2], and the minimum absolute difference of two elements is 1; 
    * for x = 2, if we apply bitwise XOR to the elements of the array with x, we get the array [4, 2, 1], and the minimum absolute difference of two elements is 1; 
    * for x = 3, if we apply bitwise XOR to the elements of the array with x, we get the array [5, 3, 0], and the minimum absolute difference of two elements is 2; 
    * for x = 4, if we apply bitwise XOR to the elements of the array with x, we get the array [2, 4, 7], and the minimum absolute difference of two elements is 2; 
    * for x = 5, if we apply bitwise XOR to the elements of the array with x, we get the array [3, 5, 6], and the minimum absolute difference of two elements is 1; 
    * for x = 6, if we apply bitwise XOR to the elements of the array with x, we get the array [0, 6, 5], and the minimum absolute difference of two elements is 1; 
    * for x = 7, if we apply bitwise XOR to the elements of the array with x, we get the array [1, 7, 4], and the minimum absolute difference of two elements is 3. 
solution: #include <bits/stdc++.h>
  using namespace std;
  struct Node {
    static const int MAX = 1 << 22;
    int low{MAX};
    int high{-MAX};
    int diff{MAX};
    operator bool() const { return low < MAX; }
  };
  void merge(vector<Node> &v, const int &start, const int &half_step) {
    for (int i = start; i < start + half_step; ++i) {
      const int only_diff = min(v[i].diff, v[i + half_step].diff);
      const int diff1 = min(only_diff, v[i + half_step].low - v[i].high);
      const int diff2 = min(only_diff, (v[i].low ^ half_step) -
                                           (v[i + half_step].high ^ half_step));
      v[i].diff = diff1;
      v[i + half_step].diff = diff2;
      const int low1 = min(v[i].low, v[i + half_step].low);
      const int low2 =
          min(v[i].low ^ half_step, v[i + half_step].low ^ half_step);
      v[i].low = low1;
      v[i + half_step].low = low2;
      const int high1 = max(v[i].high, v[i + half_step].high);
      const int high2 =
          max(v[i].high ^ half_step, v[i + half_step].high ^ half_step);
      v[i].high = high1;
      v[i + half_step].high = high2;
    }
  }
  int main() {
    int n, k;
    cin >> n >> k;
    int num = 1 << k;
    vector<Node> first_round(num);
    for (; n > 0; --n) {
      int a;
      cin >> a;
      first_round[a].low = a;
      first_round[a].high = a;
    }
    for (int deg = 0; deg < k; ++deg) {
      const int half_step = 1 << deg;
      const int step = 2 * half_step;
      for (int i = 0; i < num; i += step) merge(first_round, i, half_step);
    }
    for (int i = 0; i < num; ++i) cout << first_round[i].diff << " ";
    cout << endl;
  }
  