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>
  constexpr int inf = 1e9;
  struct Node {
    int mn, mx, sz, ans;
    Node *l, *r;
    Node(int val = 0) : ans(inf), sz(1) {
      if (val == 1) {
        mn = mx = 0;
      } else {
        mn = inf;
        mx = -inf;
      }
    }
    void recalc() {
      sz = l->sz + r->sz;
      mn = std::min(l->mn, l->sz + r->mn);
      mx = std::max(l->mx, l->sz + r->mx);
      ans = std::min({l->ans, r->ans, (l->sz + r->mn) - l->mx});
    }
    Node(Node *left, Node *right) : l(left), r(right) { recalc(); }
  };
  constexpr int K = 19;
  int n, k, cnt[1 << K];
  Node tree[2 << K];
  void swapSegments(Node *v, int d) {
    if (d == 0) {
      std::swap(v->l, v->r);
      v->recalc();
    } else {
      swapSegments(v->l, d - 1);
      swapSegments(v->r, d - 1);
      v->recalc();
    }
  }
  int ans[1 << K], cur = 0;
  void solve(int d) {
    if (d == k - 1) {
      ans[cur] = tree[1].ans;
    } else {
      solve(d + 1);
    }
    swapSegments(&tree[1], k - 1 - d);
    cur ^= 1 << d;
    if (d == k - 1) {
      ans[cur] = tree[1].ans;
    } else {
      solve(d + 1);
    }
  }
  int main() {
    std::ios::sync_with_stdio(false);
    std::cin.tie(nullptr);
    std::cin >> n >> k;
    for (int i = 0; i < n; i++) {
      int x;
      std::cin >> x;
      cnt[x]++;
    }
    for (int i = 0; i < (1 << k); i++) {
      tree[i | 1 << k] = Node(cnt[i]);
    }
    for (int i = (1 << k) - 1; i > 0; i--) {
      tree[i] = Node(&tree[2 * i], &tree[2 * i + 1]);
    }
    solve(0);
    for (int i = 0; i < (1 << k); i++) {
      std::cout << ans[i] << " \n"[i == (1 << k) - 1];
    }
    return 0;
  }
  