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 INF = int(1e9);
  const int K = 20;
  struct node {
    int max_val, min_val, ans, len;
    node(const node& left, const node& right) {
      len = left.len + right.len;
      max_val = max(left.max_val, right.max_val + left.len);
      min_val = min(left.min_val, right.min_val + left.len);
      ans = min(min(left.ans, right.ans),
                min(INF, right.min_val - left.max_val + left.len));
    }
    node(int x) {
      ans = INF;
      len = 1;
      if (x == 1)
        max_val = min_val = 0;
      else {
        max_val = -INF;
        min_val = INF;
      }
    }
    node(){};
  };
  int cnt[1 << K];
  vector<node> tree[2 << K];
  void build(int v, int l, int r) {
    tree[v].resize(r - l);
    if (l == r - 1) {
      tree[v][0] = node(cnt[l]);
    } else {
      int m = (l + r) / 2;
      build(v * 2 + 1, l, m);
      build(v * 2 + 2, m, r);
      for (int i = 0; i < m - l; i++) {
        tree[v][i] = node(tree[2 * v + 1][i], tree[2 * v + 2][i]);
        tree[v][i + (m - l)] = node(tree[2 * v + 2][i], tree[2 * v + 1][i]);
      }
    }
  }
  int main() {
    int n, k;
    scanf("%d %d", &n, &k);
    for (int i = 0; i < n; i++) {
      int x;
      scanf("%d", &x);
      cnt[x]++;
    }
    int m = 1 << k;
    build(0, 0, m);
    for (int i = 0; i < m; i++) cout << tree[0][i].ans << " ";
    return 0;
  }
  