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 k;
  struct node {
    node* c[2];
  };
  node* trie;
  void ins(int r) {
    node* f = trie;
    for (int j = k - 1; j >= 0; j--) {
      if (!f->c[1 & (r >> j)]) {
        f->c[1 & (r >> j)] = new node();
        f->c[1 & (r >> j)]->c[0] = f->c[1 & (r >> j)]->c[1] = nullptr;
      }
      f = f->c[1 & (r >> j)];
    }
  }
  vector<vector<int> > ans;
  void solve(node* l, node* r, int st, int x, int d, int diff) {
    if (d == k) {
      ans[st][x] = min(ans[st][x], diff);
    } else {
      for (int j = 0; j < 2; j++) {
        if (l->c[1 ^ j] && r->c[0 ^ j]) {
          solve(l->c[1 ^ j], r->c[0 ^ j], st, x ^ (j << (k - 1 - d)), d + 1,
                diff - (1 << (k - 1 - d)));
          continue;
        }
        if (l->c[1 ^ j] && r->c[1 ^ j]) {
          solve(l->c[1 ^ j], r->c[1 ^ j], st, x ^ (j << (k - 1 - d)), d + 1,
                diff);
          continue;
        }
        if (l->c[0 ^ j] && r->c[0 ^ j]) {
          solve(l->c[0 ^ j], r->c[0 ^ j], st, x ^ (j << (k - 1 - d)), d + 1,
                diff);
          continue;
        }
        if (l->c[0 ^ j] && r->c[1 ^ j]) {
          solve(l->c[0 ^ j], r->c[1 ^ j], st, x ^ (j << (k - 1 - d)), d + 1,
                diff + (1 << (k - 1 - d)));
          continue;
        }
      }
    }
  }
  void tr(node* v, int r, int d) {
    if (v->c[0] && v->c[1]) {
      solve(v->c[0], v->c[1], d, 0, d + 1, (1 << (k - 1 - d)));
      solve(v->c[1], v->c[0], d, (1 << (k - 1 - d)), d + 1, (1 << (k - 1 - d)));
    }
    if (v->c[0]) tr(v->c[0], 2 * r + 0, d + 1);
    if (v->c[1]) tr(v->c[1], 2 * r + 1, d + 1);
  }
  int main() {
    ios_base::sync_with_stdio(false), cin.tie(nullptr);
    trie = new node();
    trie->c[0] = trie->c[1] = nullptr;
    int n;
    cin >> n >> k;
    ans.assign(k, vector<int>(1 << k, int(1e8)));
    vector<int> a(n);
    for (int& x : a) cin >> x;
    sort(a.begin(), a.end());
    for (int x : a) ins(x);
    assert(trie != nullptr);
    tr(trie, 0, 0);
    for (int i = 0; i < (1 << k); i++) {
      int best = int(1e8);
      for (int j = 0; j < k; j++) {
        best = min(best, ans[j][i % (1 << (k - j))]);
      }
      cout << best << ' ';
    }
    cout << '\n';
  }
  