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;
  int main() {
    long long n, k;
    cin >> n >> k;
    vector<long long> act(1 << k, 0);
    for (long long i = 0; i < n; ++i) {
      long long t;
      cin >> t;
      act[t] = 1;
    }
    function<vector<vector<long long>>(long long, long long, long long)> solve =
        [&](long long l, long long r, long long ind) {
          if (l + 1 == r)
            return vector<vector<long long>>{{LLONG_MAX / 2 * (!act[l])},
                                             {-LLONG_MAX / 2 * (!act[l])},
                                             {LLONG_MAX / 2}};
          else {
            long long mid = (l + r) / 2;
            long long le = r - l;
            vector<vector<long long>> ret = {
                vector<long long>(le, LLONG_MAX / 2),
                vector<long long>(le, -LLONG_MAX / 2),
                vector<long long>(le, LLONG_MAX / 2)};
            auto q1 = solve(l, mid, ind - 1), q2 = solve(mid, r, ind - 1);
            for (long long i = 0; i < le / 2; ++i) {
              ret[0][i] = min(q1[0][i], q2[0][i] + (1 << ind));
              ret[0][i + le / 2] = min(q2[0][i], q1[0][i] + (1 << ind));
              ret[1][i + le / 2] = max(q1[1][i] + (1 << ind), q2[1][i]);
              ret[1][i] = max(q2[1][i] + (1 << ind), q1[1][i]);
              ret[2][i] = min(q1[2][i], q2[2][i]);
              ret[2][i + le / 2] = min(q1[2][i], q2[2][i]);
            }
            for (long long i = 0; i < le / 2; ++i)
              ret[2][i] = min(ret[2][i], q2[0][i] + (1 << ind) - q1[1][i]);
            for (long long i = 0; i < le / 2; ++i)
              ret[2][i + le / 2] =
                  min(ret[2][i + le / 2], q1[0][i] + (1 << ind) - q2[1][i]);
            return ret;
          }
        };
    vector<long long> ans = solve(0, 1 << k, k - 1)[2];
    for (long long i = 0; i < 1 << k; ++i) cout << ans[i] << " ";
    cout << endl;
  }
  