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>
  template <typename _Tp>
  void read(_Tp &x) {
    char ch(getchar());
    bool f(false);
    while (!isdigit(ch)) f |= ch == 45, ch = getchar();
    x = ch & 15, ch = getchar();
    while (isdigit(ch)) x = x * 10 + (ch & 15), ch = getchar();
    if (f) x = -x;
  }
  template <typename _Tp, typename... Args>
  void read(_Tp &t, Args &...args) {
    read(t);
    read(args...);
  }
  const int N = (1 << 19) + 5, inf = 1e9;
  int pool[N * 20 * 3], *P = pool;
  int *mn[N << 1], *mx[N << 1], *ans[N << 1], dep[N << 1];
  bool _[N];
  void build(int l, int r, int u) {
    mn[u] = P, P += r - l + 1, mx[u] = P, P += r - l + 1, ans[u] = P,
    P += r - l + 1;
    if (l == r) {
      if (_[l])
        mn[u][0] = 0, mx[u][0] = 0, ans[u][0] = inf;
      else
        mn[u][0] = inf, mx[u][0] = -inf, ans[u][0] = inf;
      return;
    }
    int mid = (l + r) >> 1;
    dep[((u) << 1)] = dep[u] - 1, dep[((u) << 1 | 1)] = dep[u] - 1;
    build(l, mid, ((u) << 1)), build(mid + 1, r, ((u) << 1 | 1));
    int ch[2] = {((u) << 1), ((u) << 1 | 1)}, M = (1 << (dep[u] - 1)) - 1;
    for (int i = 0; i < 1 << dep[u]; ++i) {
      int o = i >> (dep[u] - 1) & 1;
      mn[u][i] =
          std::min(mn[ch[o]][i & M], mn[ch[!o]][i & M] + (1 << (dep[u] - 1)));
      mx[u][i] =
          std::max(mx[ch[o]][i & M], mx[ch[!o]][i & M] + (1 << (dep[u] - 1)));
      ans[u][i] = std::min(
          {ans[ch[0]][i & M], ans[ch[1]][i & M],
           abs(mx[ch[o]][i & M] - (mn[ch[!o]][i & M] + (1 << (dep[u] - 1))))});
    }
  }
  int main() {
    int n, k;
    read(n, k);
    for (int i = 1, x; i <= n; ++i) read(x), _[x] = true;
    dep[1] = k, build(0, (1 << k) - 1, 1);
    for (int i = 0; i < 1 << k; ++i)
      printf("%d%c", ans[1][i], " \n"[i + 1 == (1 << k)]);
    return 0;
  }
  