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>
  int main() {
    unsigned short k;
    unsigned long n;
    std::vector<long> f, minv, maxv;
    std::ios::sync_with_stdio(false);
    std::cin.tie(nullptr);
    std::cin >> n >> k;
    minv.assign(1UL << k, 1L << k);
    maxv.assign(1UL << k, -1L << k);
    while (n--) {
      unsigned long a;
      std::cin >> a;
      minv[a] = 0;
      maxv[a] = 0;
    }
    f.assign(1UL << k, 1L << k);
    for (unsigned short i = 0; i < k; ++i) {
      for (unsigned long x = 0; x < (1UL << k); ++x) {
        if (x & (1UL << i)) continue;
        const unsigned long y = x | (1UL << i);
        f[x] = std::min(f[x], f[y]);
        f[y] = f[x];
        f[x] = std::min(f[x], minv[y] + (1L << i) - maxv[x]);
        f[y] = std::min(f[y], minv[x] + (1L << i) - maxv[y]);
        const auto minvx = minv[x], maxvx = maxv[x], minvy = minv[y],
                   maxvy = maxv[y];
        minv[x] = std::min(minvx, minvy + (1L << i));
        minv[y] = std::min(minvy, minvx + (1L << i));
        maxv[x] = std::max(maxvx, maxvy + (1L << i));
        maxv[y] = std::max(maxvy, maxvx + (1L << i));
      }
    }
    for (unsigned long i = 0; i < (1UL << k); ++i) {
      if (i) std::cout.put(' ');
      std::cout << f[i];
    }
    std::cout.put('\n');
    return 0;
  }
  