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 ll = long long;
  int const nmax = 19;
  int v[1 << nmax];
  struct Info {
    int smin;
    int pref;
    int suff;
  } dp[1 << nmax];
  int main() {
    int n, k;
    std::cin >> n >> k;
    for (int i = 1; i <= n; i++) {
      int val;
      std::cin >> val;
      v[val] = 1;
    }
    for (int i = 0; i < (1 << k); i++)
      if (v[i] == 1) {
        dp[i] = {(1 << k), 0, 0};
      } else {
        dp[i] = {(1 << k), (1 << k), (1 << k)};
      }
    for (int jump = 2; jump <= (1 << k); jump *= 2) {
      int halfjump = jump / 2;
      for (int i = 0; i < (1 << k); i += jump) {
        for (int j = i; j < i + halfjump; j++) {
          Info a = dp[j], b = dp[j + halfjump];
          dp[j].smin = std::min(std::min(a.smin, b.smin), a.suff + b.pref + 1);
          dp[j].pref = std::min(a.pref, halfjump + b.pref);
          dp[j].suff = std::min(a.suff + halfjump, b.suff);
          dp[j + halfjump].smin =
              std::min(std::min(a.smin, b.smin), a.pref + b.suff + 1);
          dp[j + halfjump].pref = std::min(b.pref, halfjump + a.pref);
          dp[j + halfjump].suff = std::min(a.suff, halfjump + b.suff);
        }
      }
    }
    for (int i = 0; i < (1 << k); i++) std::cout << dp[i].smin << " ";
    std::cout << '\n';
    return 0;
  }
  