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 M = 998244353;
  const int maxn = 1e6 + 10;
  const int inf = 1e9;
  int n, k, mx[21][1 << 19], mn[21][1 << 19], ans[21][1 << 19], sz[21][1 << 19];
  void Min(int &x, int y) {
    if (y < x) x = y;
  }
  int main() {
    scanf("%d%d", &n, &k);
    for (int i = 0; i < n; i++) {
      int x;
      scanf("%d", &x);
      sz[0][x]++;
      mx[0][x] = mn[0][x] = 0;
      ans[0][x] = inf;
    }
    for (int i = 0; i < k; i++) {
      int d = 1 << i;
      for (int j = 0; j < (1 << k); j += d * 2) {
        sz[i + 1][j] = sz[i][j] + sz[i][j + d];
        if (!sz[i + 1][j]) continue;
        if (!sz[i][j + d]) {
          for (int r = 0; r < d; r++) {
            mx[i + 1][j + r] = mx[i][j + r];
            mn[i + 1][j + r] = mn[i][j + r];
            mx[i + 1][j + r + d] = mx[i][j + r] + d;
            mn[i + 1][j + r + d] = mn[i][j + r] + d;
            ans[i + 1][j + r] = ans[i + 1][j + r + d] = ans[i][j + r];
          }
        } else if (!sz[i][j]) {
          for (int r = 0; r < d; r++) {
            mx[i + 1][j + r] = mx[i][j + r + d] + d;
            mn[i + 1][j + r] = mn[i][j + r + d] + d;
            mx[i + 1][j + r + d] = mx[i][j + r + d];
            mn[i + 1][j + r + d] = mn[i][j + r + d];
            ans[i + 1][j + r] = ans[i + 1][j + r + d] = ans[i][j + r + d];
          }
        } else {
          for (int r = 0; r < d; r++) {
            mx[i + 1][j + r] = mx[i][j + r + d] + d;
            mn[i + 1][j + r] = mn[i][j + r];
            mx[i + 1][j + r + d] = mx[i][j + r] + d;
            mn[i + 1][j + r + d] = mn[i][j + r + d];
            ans[i + 1][j + r] = ans[i + 1][j + r + d] =
                min(ans[i][j + r], ans[i][j + r + d]);
            Min(ans[i + 1][j + r], (mn[i][j + r + d] + d) - mx[i][j + r]);
            Min(ans[i + 1][j + r + d], (mn[i][j + r] + d) - mx[i][j + r + d]);
          }
        }
      }
    }
    for (int i = 0; i < (1 << k); i++) printf("%d ", ans[k][i]);
    puts("");
    return 0;
  }
  