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;
  inline long long read() {
    char c = getchar();
    long long first = 0;
    bool f = 0;
    for (; !isdigit(c); c = getchar()) f ^= !(c ^ 45);
    for (; isdigit(c); c = getchar())
      first = (first << 1) + (first << 3) + (c ^ 48);
    if (f) first = -first;
    return first;
  }
  const long long inf = 1e18;
  const long long N = 20;
  const long long M = (1 << 19) + 10;
  long long k, n, a[M], f[N][M];
  long long son[M << 1][2], cnt;
  vector<long long> minn[M << 1], maxn[M << 1];
  void insert(long long first) {
    long long now = 0;
    for (long long i = k - 1; i >= 0; i--) {
      long long c = ((first >> i) & 1);
      if (!son[now][c]) son[now][c] = ++cnt;
      now = son[now][c];
    }
  }
  void dfs(long long u, long long first) {
    if (first == 0) {
      minn[u].push_back(0), maxn[u].push_back(0);
      return;
    }
    if (son[u][0]) dfs(son[u][0], first - 1);
    if (son[u][1]) dfs(son[u][1], first - 1);
    for (long long j = 0; j < (1 << first); j++) {
      long long c = (j >> (first - 1)) & 1, second = j - (1 << (first - 1)) * c;
      long long Min = inf, Max = 0;
      if (son[u][c])
        Min = min(Min, minn[son[u][c]][second]),
        Max = max(Max, maxn[son[u][c]][second]);
      if (son[u][c ^ 1])
        Min = min(Min, minn[son[u][c ^ 1]][second] + (1 << (first - 1))),
        Max = max(Max, maxn[son[u][c ^ 1]][second] + (1 << (first - 1)));
      if (son[u][c] && son[u][c ^ 1])
        f[first][j] =
            min(f[first][j], (1 << (first - 1)) + minn[son[u][c ^ 1]][second] -
                                 maxn[son[u][c]][second]);
      minn[u].push_back(Min), maxn[u].push_back(Max);
    }
  }
  signed main() {
    n = read(), k = read();
    for (long long i = 1; i <= n; i++) a[i] = read(), insert(a[i]);
    memset(f, 0x3f, sizeof(f));
    dfs(0, k);
    for (long long i = 0; i < (1 << k); i++) {
      long long ans = inf;
      for (long long d = 1; d <= k; d++) ans = min(ans, f[d][i & ((1 << d) - 1)]);
      cout << ans << ' ';
    }
    return 0;
  }
  