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 n, k;
  int a[1 << 20];
  int ch[1 << 20][2], maxi[1 << 20], mini[1 << 20], ans[1 << 20], fa[1 << 20],
      prefix[1 << 20], dep[1 << 20];
  int output[1 << 20];
  vector<int> multi[1 << 20];
  int root = 1, cnt = 0;
  void all_init() {
    memset(mini, 0x3f, sizeof(mini));
    memset(maxi, -0x3f, sizeof(maxi));
    memset(ans, 0x3f, sizeof(ans));
  }
  void upgrade(int k, int c, int x) {
    if (k == c) return;
    upgrade(k - 1, c, ch[x][0]);
    upgrade(k - 1, c, ch[x][1]);
    ans[x] =
        min(mini[ch[x][1]] + prefix[ch[x][1]] - maxi[ch[x][0]] - prefix[ch[x][0]],
            min(ans[ch[x][0]], ans[ch[x][1]]));
    maxi[x] = max(maxi[ch[x][0]] + prefix[ch[x][0]] - prefix[x],
                  maxi[ch[x][1]] + prefix[ch[x][1]] - prefix[x]);
    mini[x] = min(mini[ch[x][0]] + prefix[ch[x][0]] - prefix[x],
                  mini[ch[x][1]] + prefix[ch[x][1]] - prefix[x]);
  }
  int build(int d, int limit) {
    if (d < 0) return 0;
    int x = ++cnt;
    multi[d].push_back(x);
    prefix[x] = limit;
    dep[x] = d;
    ch[x][0] = build(d - 1, limit);
    ch[x][1] = build(d - 1, limit | (1 << (d - 1)));
    fa[ch[x][0]] = fa[ch[x][1]] = x;
    return x;
  }
  void insert(int v) {
    int x = root;
    for (int i = k - 1; i >= 0; --i) {
      int s = v >> i & 1;
      x = ch[x][s];
      maxi[x] = max(maxi[x], v - prefix[x]);
      mini[x] = min(mini[x], v - prefix[x]);
    }
  }
  void flip(int c) {
    for (int i = 0; i < multi[c].size(); ++i) {
      int x = multi[c][i];
      swap(ch[x][0], ch[x][1]);
      swap(prefix[ch[x][0]], prefix[ch[x][1]]);
    }
    upgrade(k, c - 1, root);
  }
  void DFS(int d, int x) {
    if (d == k) {
      output[x] = ans[root];
      return;
    }
    DFS(d + 1, x);
    flip(d + 1);
    DFS(d + 1, x | (1 << d));
    flip(d + 1);
  }
  int main() {
    all_init();
    scanf("%d%d", &n, &k);
    build(k, 0);
    fa[0] = 0;
    for (int i = 1; i <= n; ++i) {
      scanf("%d", &a[i]);
      insert(a[i]);
    }
    upgrade(k, 0, root);
    DFS(0, 0);
    for (int i = 0; i < (1 << k); ++i) printf("%d ", output[i]);
  }
  