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 MK = 20, MN = 1 << MK, INF = 0x3f3f3f3f;
  int N, K, M;
  struct node {
    int mn, mx, ans;
    node(int mn = -1, int mx = -1, int ans = INF) : mn(mn), mx(mx), ans(ans) {}
    operator bool() { return mn > -1; }
  } tree[MK][MN];
  int main() {
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    cin >> N >> K;
    M = 1 << K;
    for (int i = 0, x; i < N; i++) {
      cin >> x;
      tree[0][x] = {x, x, INF};
    }
    for (int lv = 1, sz, psz; sz = 1 << lv, psz = sz >> 1, lv <= K; lv++) {
      for (int i = 0; i < M; i += sz) {
        for (int bit = 0; bit < psz; ++bit) {
          node a = tree[lv - 1][i | bit], b = tree[lv - 1][i | psz | bit],
               &cur0 = tree[lv][i | bit], &cur1 = tree[lv][i | psz | bit];
          if (a && b) {
            cur0 = {a.mn, b.mx, min(b.mn - a.mx, min(a.ans, b.ans))};
            cur1 = {b.mn ^ psz, a.mx ^ psz,
                    min(sz - b.mx + a.mn, min(a.ans, b.ans))};
          } else {
            cur0 = a ? a : b;
            cur1 = cur0 ? node(cur0.mn ^ psz, cur0.mx ^ psz, cur0.ans) : cur0;
          }
        }
      }
    }
    for (int bit = 0; bit < M; ++bit) {
      printf("%d%c", tree[K][bit].ans, " \n"[bit == M - 1]);
    }
    return 0;
  }
  