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>
  const int INF = 0x3f3f3f3f;
  struct state {
    int mn, mx, s;
    state() : mn(INF), mx(-INF), s(INF) {}
    state(int v) : mn(v), mx(v), s(INF) {}
    state(int _mn, int _mx, int _s) : mn(_mn), mx(_mx), s(_s) {}
    state operator+(const state &rhs) const {
      return state(std::min(mn, rhs.mn), std::max(mx, rhs.mx),
                   std::min({s, rhs.s, rhs.mn - mx}));
    }
  };
  int main() {
    std::ios_base::sync_with_stdio(false);
    std::cin.tie(0);
    int n, m;
    std::cin >> m >> n;
    std::vector<bool> vis(1 << n);
    for (int i = 0; i < m; ++i) {
      int x;
      std::cin >> x;
      vis[x] = true;
    }
    std::vector<state> f(1 << n);
    for (int x = 0; x < (1 << n); ++x) {
      if (vis[x]) {
        f[x] = 0;
      }
    }
    for (int i = 0; i < n; ++i) {
      std::vector<state> g(1 << n);
      for (int x = 0; x < (1 << n); ++x) {
        state a = f[x], b = f[x ^ (1 << i)];
        if (b.mn != -INF) {
          b.mn ^= 1 << i;
          b.mx ^= 1 << i;
        }
        g[x] = a + b;
      }
      f.swap(g);
    }
    for (int x = 0; x < (1 << n); ++x) {
      std::cout << f[x].s << " ";
    }
    std::cout << "\n";
  }
  