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 store[20][1 << 20];
  struct trie {
    int k;
    struct Node {
      Node* ch[2];
      Node() { ch[0] = ch[1] = NULL; }
    };
    Node* root;
    trie(int k) : k(k) { root = new Node(); }
    void insert(int x) {
      Node* cur = root;
      for (int i = k - 1; i >= 0; i--) {
        int side = x >> i & 1;
        if (!cur->ch[side]) cur->ch[side] = new Node();
        cur = cur->ch[side];
      }
    }
    void eval() {
      vector<int> ret(1 << k);
      for (int i = 0; i <= k; i++) fill(store[i], store[i] + (1 << k), 1 << 30);
      function<vector<pair<int, int> >(Node*, int)> dfs = [&](Node* node, int h) {
        if (!node) return vector<pair<int, int> >();
        if (h == 0) {
          vector<pair<int, int> > ret = {{0, 0}};
          return ret;
        }
        vector<pair<int, int> > A = dfs(node->ch[0], h - 1),
                                B = dfs(node->ch[1], h - 1);
        vector<pair<int, int> > ret(1 << h);
        for (int x = 0; x < (1 << h); x++) {
          int add = 1 << (h - 1);
          int rem = x & (add - 1);
          if (x >> (h - 1) & 1) {
            ret[x].first = node->ch[1] ? B[rem].first : A[rem].first + add;
            ret[x].second = node->ch[0] ? A[rem].second + add : B[rem].second;
          } else {
            ret[x].first = node->ch[0] ? A[rem].first : B[rem].first + add;
            ret[x].second = node->ch[1] ? B[rem].second + add : A[rem].second;
          }
          if (!A.empty() && !B.empty()) {
            if (x >> (h - 1) & 1)
              store[h][x] = min(store[h][x], A[rem].first + add - B[rem].second);
            else
              store[h][x] = min(store[h][x], B[rem].first + add - A[rem].second);
          }
          assert(max(ret[x].first, ret[x].second) < (1 << h) &&
                 min(ret[x].first, ret[x].second) >= 0);
        }
        return ret;
      };
      dfs(root, k);
    }
  };
  int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    int n, k;
    cin >> n >> k;
    trie T(k);
    for (int i = 0; i < n; i++) {
      int x;
      cin >> x;
      T.insert(x);
    }
    T.eval();
    for (int x = 0; x < (1 << k); x++) {
      int ans = 1 << 30;
      for (int h = k; h; h--) {
        ans = min(ans, store[h][x & ((1 << h) - 1)]);
      }
      cout << ans << " ";
    }
    cout << '\n';
  }
  