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;
  const int inf = 1e9;
  int mx[1 << 19], mn[1 << 19], f[1 << 19];
  signed main() {
    ios::sync_with_stdio(false);
    int n, k;
    cin >> n >> k;
    for (int i = 0; i < (1 << k); ++i) mx[i] = -inf, mn[i] = f[i] = inf;
    for (int i = 1; i <= n; ++i) {
      int x;
      cin >> x;
      mx[x] = mn[x] = 0;
    }
    for (int d = 0; d < k; ++d)
      for (int i = 0; i < (1 << k); ++i)
        if (i >> d & 1) {
          int l = i ^ (1 << d), r = i;
          int x = f[l], y = f[r];
          f[l] = min({f[l], y, mn[r] - mx[l] + (1 << d)}),
          f[r] = min({f[r], x, mn[l] - mx[r] + (1 << d)});
          int mxl = mx[l], mxr = mx[r], mnl = mn[l], mnr = mn[r];
          mx[l] = max(mx[l], mxr + (1 << d));
          mx[r] = max(mx[r], mxl + (1 << d));
          mn[l] = min(mn[l], mnr + (1 << d));
          mn[r] = min(mn[r], mnl + (1 << d));
        }
    for (int i = 0; i < (1 << k); ++i) cout << f[i] << " ";
  }
  