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;
  int n, k, f[1 << 19], mx[1 << 19], mn[1 << 19];
  const int inf = 1 << 25;
  int main() {
    scanf("%d%d", &n, &k);
    for (int i = 0; i < 1 << k; i++) f[i] = mn[i] = inf, mx[i] = -inf;
    for (int i = 0; i < n; i++) {
      int x;
      scanf("%d", &x);
      mx[x] = mn[x] = 0;
    }
    for (int i = 0; i < k; i++)
      for (int x = 0; x < 1 << k; x++)
        if (x & (1 << i)) {
          int y = x ^ (1 << i);
          f[x] = f[y] = min(f[x], f[y]);
          f[x] = min(f[x], mn[y] - mx[x] + (1 << i));
          f[y] = min(f[y], mn[x] - mx[y] + (1 << i));
          int mnx = mn[x], mny = mn[y], mxx = mx[x], mxy = mx[y];
          mn[x] = min(mnx, mny + (1 << i));
          mn[y] = min(mny, mnx + (1 << i));
          mx[x] = max(mxx, mxy + (1 << i));
          mx[y] = max(mxy, mxx + (1 << i));
        }
    for (int i = 0; i < 1 << k; i++) printf("%d ", f[i]);
  }
  