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 a[1 << 20];
  struct Node {
    Node* ch[2];
    int ans;
    int mi1, mi2;
  } node[1 << 21];
  int ans[1 << 20];
  int tot = 0;
  void pushUp(Node* rt, int level) {
    rt->mi1 = 1e9;
    rt->mi2 = 1e9;
    rt->ans = 1e9;
    if (rt->ch[0])
      rt->mi1 = rt->ch[0]->mi1;
    else
      rt->mi1 = (1 << level) + rt->ch[1]->mi1;
    if (rt->ch[1])
      rt->mi2 = rt->ch[1]->mi2;
    else
      rt->mi2 = rt->ch[0]->mi2 + (1 << level);
    if (rt->ch[0]) rt->ans = min(rt->ans, rt->ch[0]->ans);
    if (rt->ch[1]) rt->ans = min(rt->ans, rt->ch[1]->ans);
    if (rt->ch[0] && rt->ch[1])
      rt->ans = min(rt->ans, rt->ch[1]->mi1 + rt->ch[0]->mi2 + 1);
  }
  void flip(int cur, int level, Node* rt) {
    if (rt == NULL) return;
    if (cur == level) {
      swap(rt->ch[0], rt->ch[1]);
      pushUp(rt, level);
    } else {
      flip(cur - 1, level, rt->ch[0]);
      flip(cur - 1, level, rt->ch[1]);
      pushUp(rt, cur);
    }
  }
  void dfs(int cur, int v, int k, Node* rt) {
    if (cur == k) {
      ans[v] = rt->ans;
      return;
    }
    dfs(cur + 1, v, k, rt);
    flip(k - 1, cur, rt);
    dfs(cur + 1, v | (1 << cur), k, rt);
    flip(k - 1, cur, rt);
  }
  Node* newNode() {
    Node* rt = &node[tot++];
    rt->ch[0] = rt->ch[1] = NULL;
    rt->ans = 1e9;
    rt->mi1 = 1e9;
    rt->mi2 = 1e9;
    return rt;
  }
  void insert(int cur, int v, Node* rt) {
    if (cur < 0) {
      rt->mi1 = rt->mi2 = 0;
      rt->ans = 1e9;
      return;
    }
    if (v & (1 << cur)) {
      if (rt->ch[1] == NULL) rt->ch[1] = newNode();
      insert(cur - 1, v, rt->ch[1]);
    } else {
      if (rt->ch[0] == NULL) rt->ch[0] = newNode();
      insert(cur - 1, v, rt->ch[0]);
    }
    pushUp(rt, cur);
  }
  int main(void) {
    int n, k;
    scanf("%d %d", &n, &k);
    for (int i = 0; i < n; i++) scanf("%d", &a[i]);
    Node* rt = newNode();
    for (int i = 0; i < n; i++) {
      insert(k - 1, a[i], rt);
    }
    dfs(0, 0, k, rt);
    for (int i = 0; i < (1 << k); i++) printf("%d ", ans[i]);
    puts("");
    return 0;
  }
  