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 nax = (1 << 19);
  int a[nax];
  int n, k;
  struct node {
    node *lft, *rgt;
    int prefix, len;
    vector<int> mini, maxi, ans, numbers;
  };
  node* root;
  void dfs(node* cur) {
    if (cur->len == k) {
      cur->mini.resize(1);
      cur->maxi.resize(1);
      cur->ans.resize(1);
      cur->mini[0] = 0;
      cur->maxi[0] = 0;
      cur->ans[0] = 1e9;
      return;
    }
    if (cur->len > k) return;
    if (cur->numbers.empty()) return;
    cur->rgt = new node;
    cur->lft = new node;
    for (int x : cur->numbers) {
      int v = (x & (1 << (k - cur->len - 1)));
      if (v) {
        cur->rgt->numbers.push_back(x ^ v);
      } else
        cur->lft->numbers.push_back(x ^ v);
    }
    cur->lft->len = cur->rgt->len = cur->len + 1;
    cur->lft->prefix = cur->prefix * 2;
    cur->rgt->prefix = cur->prefix * 2 + 1;
    dfs(cur->lft);
    dfs(cur->rgt);
    int sz = (1 << (k - cur->len));
    cur->mini.resize(sz, 1e9);
    cur->maxi.resize(sz, -1e9);
    cur->ans.resize(sz, 1e9);
    if (cur->lft->numbers.size()) {
      for (int i = 0; i < sz / 2; i++) {
        cur->ans[i] = min(cur->ans[i], cur->lft->ans[i]);
        cur->ans[i + sz / 2] = min(cur->ans[i + sz / 2], cur->lft->ans[i]);
      }
    }
    if (cur->rgt->numbers.size()) {
      for (int i = 0; i < sz / 2; i++) {
        cur->ans[i] = min(cur->ans[i], cur->rgt->ans[i]);
        cur->ans[i + sz / 2] = min(cur->ans[i + sz / 2], cur->rgt->ans[i]);
      }
    }
    int ful = (sz / 2 - 1);
    for (int i = 0; i < sz; i++) {
      int B = (i & (sz / 2));
      if (cur->lft->numbers.size()) {
        cur->mini[i] = min(cur->mini[i], B + cur->lft->mini[i & ful]);
        cur->maxi[i] = max(cur->maxi[i], B + cur->lft->maxi[i & ful]);
      }
      if (cur->rgt->numbers.size()) {
        cur->mini[i] = min(cur->mini[i], sz / 2 - B + cur->rgt->mini[i & ful]);
        cur->maxi[i] = max(cur->maxi[i], sz / 2 - B + cur->rgt->maxi[i & ful]);
      }
      if (cur->lft->numbers.size() && cur->rgt->numbers.size()) {
        if (B == 0)
          cur->ans[i] = min(cur->ans[i], cur->rgt->mini[i & ful] + sz / 2 -
                                             cur->lft->maxi[i & ful]);
        else
          cur->ans[i] = min(cur->ans[i], cur->lft->mini[i & ful] + sz / 2 -
                                             cur->rgt->maxi[i & ful]);
      }
    }
  }
  void solve() {
    cin >> n >> k;
    for (int i = 0; i < n; i++) cin >> a[i];
    root = new node;
    root->prefix = 0;
    root->len = 0;
    root->numbers.resize(n);
    for (int i = 0; i < n; i++) root->numbers[i] = a[i];
    dfs(root);
    for (int i = 0; i < (1 << k); i++) cout << root->ans[i] << " ";
  }
  int main() {
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    int tt = 1;
    while (tt--) solve();
    return 0;
  }
  