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 maxn = 1111111, mod = 998244353;
  template <typename T>
  inline void read(T &x) {
    x = 0;
    bool f = 0;
    char ch = getchar();
    while (ch < '0' || ch > '9') f |= ch == '-', ch = getchar();
    while (ch >= '0' && ch <= '9') x = x * 10 + ch - '0', ch = getchar();
    if (f) x = -x;
  }
  int n, k, a[maxn], rt, cnt, ch[maxn * 20][2];
  void insert(int &x, int v, int dep) {
    if (!x) x = ++cnt;
    if (dep == -1) return;
    insert(ch[x][(v >> dep) & 1], v, dep - 1);
  }
  struct wtf {
    vector<int> s, mx, mn;
  };
  wtf dfs(int x, int dep) {
    if (dep == -1) {
      wtf w;
      w.s = vector<int>(1, 1e9);
      w.mx = w.mn = vector<int>(1, 0);
      return w;
    }
    int lim = (1 << dep);
    vector<int> s(lim * 2, 1e9), mx(lim * 2, 0), mn(lim * 2, 1e9);
    wtf l, r;
    for (int _ = (0); _ <= (1); _++)
      if (ch[x][_]) {
        wtf w = dfs(ch[x][_], dep - 1);
        if (_)
          r = w;
        else
          l = w;
        for (int i = (0); i <= (lim - 1); i++) {
          s[i] = min(s[i], w.s[i]);
          s[i + lim] = min(s[i + lim], w.s[i]);
          mx[i] = max(mx[i], w.mx[i] ^ (_ * lim));
          mx[i + lim] = max(mx[i + lim], w.mx[i] ^ ((_ ^ 1) * lim));
          mn[i] = min(mn[i], w.mn[i] ^ (_ * lim));
          mn[i + lim] = min(mn[i + lim], w.mn[i] ^ ((_ ^ 1) * lim));
        }
      }
    if (ch[x][0] && ch[x][1]) {
      for (int i = (0); i <= (lim - 1); i++)
        s[i] = min(s[i], r.mn[i] - l.mx[i] + lim);
      for (int i = (0); i <= (lim - 1); i++)
        s[i + lim] = min(s[i + lim], l.mn[i] - r.mx[i] + lim);
    }
    return (wtf){s, mx, mn};
  }
  int main() {
    read(n);
    read(k);
    for (int i = (1); i <= (n); i++) read(a[i]);
    for (int i = (1); i <= (n); i++) insert(rt, a[i], k - 1);
    vector<int> ans = dfs(rt, k - 1).s;
    for (int i = (0); i <= ((1 << k) - 1); i++) printf("%d ", ans[i]);
  }
  