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;
  using ll = long long;
  using ld = long double;
  const int INF = 1e9;
  struct Node {
    Node* ch[2];
    int mn = INF;
    int mx = -INF;
    int minAdj = INF;
    void pull(int l) {
      mn = min(ch[0]->mn, ch[1]->mn + (1 << l));
      mx = max(ch[0]->mx, ch[1]->mx + (1 << l));
      minAdj = min(ch[0]->minAdj, ch[1]->minAdj);
      minAdj = min(minAdj, ch[1]->mn + (1 << l) - ch[0]->mx);
    }
  };
  Node* newNode() {
    static deque<Node> cache;
    cache.push_back(Node());
    return &cache.back();
  }
  vector<bool> has;
  void init(Node* x, int l, int a) {
    if (l < 0) {
      if (has[a]) x->mn = x->mx = 0;
      return;
    }
    for (int i = 0; i < 2; ++i) {
      x->ch[i] = newNode();
      init(x->ch[i], l - 1, a * 2 + i);
    }
    x->pull(l);
  }
  void toggle(Node* x, int l, int target) {
    if (l == target) {
      swap(x->ch[0], x->ch[1]);
    } else {
      toggle(x->ch[0], l - 1, target);
      toggle(x->ch[1], l - 1, target);
    }
    x->pull(l);
  }
  vector<int> genGrayCode(int n) {
    if (n == 1) return {0, 1};
    auto ret = genGrayCode(n - 1);
    ret.resize(1 << n);
    for (int i = 0; i < (1 << (n - 1)); ++i)
      ret[(1 << n) - i - 1] = ret[i] | (1 << (n - 1));
    return ret;
  }
  int main() {
    int n, k;
    scanf("%d%d", &n, &k);
    has.assign(1 << k, false);
    for (int i = 0; i < n; ++i) {
      int a;
      scanf("%d", &a);
      has[a] = true;
    }
    Node* root = newNode();
    init(root, k - 1, 0);
    auto g = genGrayCode(k);
    vector<int> revG(g.size());
    for (int i = 0; i < g.size(); ++i) {
      for (int j = 0; j < k; ++j)
        if (g[i] & (1 << j)) revG[i] |= 1 << (k - j - 1);
    }
    vector<int> ans(1 << k);
    ans[0] = root->minAdj;
    for (int i = 1; i < revG.size(); ++i) {
      int v = revG[i] ^ revG[i - 1];
      int bit = 31 - __builtin_clz(v);
      toggle(root, k - 1, bit);
      ans[revG[i]] = root->minAdj;
    }
    for (int a : ans) printf("%d ", a);
  }
  