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 mod = 998244353;
  const int K = 19;
  const int K_N = (1 << (K + 1));
  vector<vector<int> > nodes[K_N];
  int len[K_N];
  int have[K_N];
  void go_to(int v, int s, int i, int k) {
    if (i == k) {
      if (have[s]) {
        nodes[v].push_back({0, 0, K_N});
      } else {
        nodes[v].push_back({-1, -1, -1});
      }
      return;
    } else {
      go_to(v * 2, s * 2, i + 1, k);
      go_to(v * 2 + 1, s * 2 + 1, i + 1, k);
      int t1 = (1 << (k - i - 1));
      for (int u = 0; u < (1 << (k - i)); u++) {
        int u1 = u & ((1 << (k - i - 1)) - 1);
        if (nodes[v * 2][u1][0] == -1 && nodes[v * 2 + 1][u1][0] == -1) {
          nodes[v].push_back({-1, -1, -1});
        } else if (nodes[v * 2][u1][0] == -1) {
          if (u & t1) {
            nodes[v].push_back(nodes[v * 2 + 1][u1]);
          } else {
            nodes[v].push_back({nodes[v * 2 + 1][u1][0] + t1,
                                nodes[v * 2 + 1][u1][1] + t1,
                                nodes[v * 2 + 1][u1][2]});
          }
        } else if (nodes[v * 2 + 1][u1][0] == -1) {
          if (u & t1) {
            nodes[v].push_back({nodes[v * 2][u1][0] + t1,
                                nodes[v * 2][u1][1] + t1, nodes[v * 2][u1][2]});
          } else {
            nodes[v].push_back(nodes[v * 2][u1]);
          }
        } else {
          if (u & (1 << (k - i - 1))) {
            int d1 = min(nodes[v * 2][u1][2], nodes[v * 2 + 1][u1][2]);
            d1 = min(d1, nodes[v * 2][u1][0] + t1 - nodes[v * 2 + 1][u1][1]);
            nodes[v].push_back(
                {nodes[v * 2 + 1][u1][0], nodes[v * 2][u1][1] + t1, d1});
          } else {
            int d1 = min(nodes[v * 2][u1][2], nodes[v * 2 + 1][u1][2]);
            d1 = min(d1, nodes[v * 2 + 1][u1][0] + t1 - nodes[v * 2][u1][1]);
            nodes[v].push_back(
                {nodes[v * 2][u1][0], nodes[v * 2 + 1][u1][1] + t1, d1});
          }
        }
      }
    }
  }
  signed main() {
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    int n, k;
    cin >> n >> k;
    for (int i = 0; i < n; i++) {
      int x;
      cin >> x;
      have[x] = 1;
    }
    go_to(1, 0, 0, k);
    for (int i = 0; i < (1 << k); i++) {
      cout << nodes[1][i][2] << " ";
    }
    return 0;
  }
  