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>
  inline void chkmin(int &x, int y) { x = std::min(x, y); }
  const int MK = 19;
  const int MN = 1 << MK | 7;
  int N, K, A[MN];
  int ch[MN * MK][2], cnt;
  inline void Insert(int x) {
    int now = 1;
    for (int j = K - 1; j >= 0; --j) {
      int y = x >> j & 1;
      if (!ch[now][y]) ch[now][y] = ++cnt;
      now = ch[now][y];
    }
  }
  int f[MK][1 << MK];
  int Minimum(int u, int j, int x) {
    if (j == -1) return 0;
    int d = x >> j & 1;
    if (ch[u][d]) return Minimum(ch[u][d], j - 1, x);
    return Minimum(ch[u][d ^ 1], j - 1, x) | 1 << j;
  }
  int Maximum(int u, int j, int x) {
    if (j == -1) return 0;
    int d = x >> j & 1;
    if (ch[u][d ^ 1]) return Maximum(ch[u][d ^ 1], j - 1, x) | 1 << j;
    return Maximum(ch[u][d], j - 1, x);
  }
  void DFS(int u, int j) {
    if (ch[u][0] && ch[u][1]) {
      for (int x = 0; x < 1 << j; ++x) {
        chkmin(f[j][x], Minimum(ch[u][1], j - 1, x) + (1 << j) -
                            Maximum(ch[u][0], j - 1, x));
        chkmin(f[j][x | 1 << j], Minimum(ch[u][0], j - 1, x) + (1 << j) -
                                     Maximum(ch[u][1], j - 1, x));
      }
    }
    if (ch[u][0]) DFS(ch[u][0], j - 1);
    if (ch[u][1]) DFS(ch[u][1], j - 1);
  }
  int Ans[MN];
  int main() {
    cnt = 1;
    scanf("%d%d", &N, &K);
    for (int i = 1; i <= N; ++i) scanf("%d", &A[i]), Insert(A[i]);
    for (int j = 0; j < K; ++j)
      for (int i = 0; i < 2 << j; ++i) f[j][i] = 1 << K;
    DFS(1, K - 1);
    for (int i = 0; i < 1 << K; ++i) Ans[i] = 1 << K;
    for (int j = 0; j < K; ++j) {
      for (int i = 0; i < 2 << j; ++i) {
        for (int p = 0; p < 1 << (K - j - 1); ++p)
          chkmin(Ans[p << (j + 1) | i], f[j][i]);
      }
    }
    for (int i = 0; i < 1 << K; ++i)
      printf("%d%c", Ans[i], " \n"[i == (1 << K) - 1]);
    return 0;
  }
  