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 kmax = 19;
  int seen[kmax][(1 << kmax) + 42];
  int n, k;
  int noted[kmax][(1 << kmax) + 42];
  int main() {
    scanf("%i%i", &n, &k);
    for (int b = 0; b < k; b++)
      for (int i = 0; i < (1 << k); i++) noted[b][i] = 1 << k;
    for (int i = 1; i <= n; i++) {
      int val;
      scanf("%i", &val);
      for (int j = 0; j < k; j++) {
        seen[j][val]++;
        val = val / 2;
      }
    }
    for (int diff = 0; diff < k; diff++) {
      for (int a_high = 0; a_high < (1 << (k - 1 - diff)); a_high++)
        for (int x = 0; x < (1 << (diff + 1)); x++) {
          int low = a_high;
          int high = a_high;
          if ((x & (1 << diff)))
            low = low * 2 + 1, high = high * 2;
          else
            low = low * 2, high = high * 2 + 1;
          if (seen[diff][low] == 0 || seen[diff][high] == 0) continue;
          for (int j = diff - 1; j >= 0; j--) {
            int has = 0;
            if ((x & (1 << j))) has = 1;
            if (seen[j][low * 2 + (!has)] && seen[j][high * 2 + has])
              low = low * 2 + (!has), high = high * 2 + has;
            else if (seen[j][low * 2 + has] && seen[j][high * 2 + has])
              low = low * 2 + has, high = high * 2 + has;
            else if (seen[j][low * 2 + (!has)] && seen[j][high * 2 + (!has)])
              low = low * 2 + (!has), high = high * 2 + (!has);
            else
              low = low * 2 + has, high = high * 2 + (!has);
          }
          noted[diff][x] = min(noted[diff][x], (high ^ x) - (low ^ x));
        }
    }
    for (int x = 0; x < (1 << k); x++) {
      int outp = 1 << k;
      for (int j = k - 1; j >= 0; j--) {
        int mod = (1 << (j + 1)) - 1;
        int remain = x & mod;
        outp = min(outp, noted[j][remain]);
      }
      printf("%i ", outp);
    }
    printf("\n");
    return 0;
  }
  