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 int64 = long long;
  const int N = (1 << 19) + 5;
  int n, k, a[N];
  struct Node {
    int min, max, ans;
  };
  Node f[2][N];
  void work() {
    scanf("%d%d", &n, &k);
    for (int i = 0; i < n; ++i) {
      int x;
      scanf("%d", &x);
      a[x] = 1;
    }
    int cur = 0, nxt = 1;
    for (int i = 0; i < 1 << k; ++i) {
      f[cur][i].ans = N;
      if (a[i]) {
        f[cur][i].min = i;
        f[cur][i].max = i;
      } else {
        f[cur][i].min = N;
        f[cur][i].max = -N;
      }
    }
    for (int j = 0; j < k; ++j) {
      for (int i = 0; i < 1 << k; ++i) {
        Node left = f[cur][i], right = f[cur][i ^ (1 << j)];
        if (i >> j & 1) {
          if (left.min < N) {
            left.min -= 1 << j;
            left.max -= 1 << j;
          }
          if (right.min < N) {
            right.min += 1 << j;
            right.max += 1 << j;
          }
        }
        f[nxt][i].min = std::min(left.min, right.min);
        f[nxt][i].max = std::max(left.max, right.max);
        f[nxt][i].ans = std::min(left.ans, right.ans);
        if (right.min < N && left.max >= 0) {
          f[nxt][i].ans = std::min(f[nxt][i].ans, right.min - left.max);
        }
      }
      std::swap(cur, nxt);
    }
    for (int i = 0; i < 1 << k; ++i) {
      printf("%d%c", f[cur][i].ans, " \n"[i + 1 == (1 << k)]);
    }
  }
  int main() {
    int T = 1;
    while (T--) {
      work();
    }
    return 0;
  }
  