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;
  char _c;
  bool _f;
  template <class T>
  inline void IN(T& x) {
    x = 0, _f = 0;
    while (_c = getchar(), !isdigit(_c))
      if (_c == '-') _f = 1;
    while (isdigit(_c)) x = x * 10 + _c - '0', _c = getchar();
    if (_f) x = -x;
  }
  template <class T>
  inline void chkmin(T& x, T y) {
    if (x > y) x = y;
  }
  template <class T>
  inline void chkmax(T& x, T y) {
    if (x < y) x = y;
  }
  const int N = (1 << 19) | 5;
  const int M = N * 20;
  const int inf = 1e9 + 7;
  int n, k, a[N];
  int cnt, mi[M], mx[M], dep[M], siz[M], ans[M], ch[M][2];
  inline void pushup(int now) {
    mi[now] = ch[now][0] ? mi[ch[now][0]] : (1 << dep[now]) + mi[ch[now][1]];
    mx[now] = ch[now][1] ? (1 << dep[now]) + mx[ch[now][1]] : mx[ch[now][0]];
    if (siz[now] <= 1)
      ans[now] = inf;
    else {
      if (!ch[now][0] || !ch[now][1])
        ans[now] = ans[ch[now][ch[now][0] == 0]];
      else
        ans[now] = min(min(ans[ch[now][0]], ans[ch[now][1]]),
                       mi[ch[now][1]] + (1 << dep[now]) - mx[ch[now][0]]);
    }
  }
  void insert(int now, int val) {
    ++siz[now];
    if (dep[now] == -1) return ans[now] = inf, void();
    bool c = (val >> dep[now]) & 1;
    if (!ch[now][c]) ch[now][c] = ++cnt, dep[ch[now][c]] = dep[now] - 1;
    insert(ch[now][c], val), pushup(now);
  }
  void flip(int now, int val) {
    if (dep[now] == -1 || (!((val >> dep[now]) & 1))) return;
    swap(ch[now][0], ch[now][1]);
    if (ch[now][0]) flip(ch[now][0], val);
    if (ch[now][1]) flip(ch[now][1], val);
    pushup(now);
  }
  int ANS[N], lst = 0;
  void solve(int val, int now) {
    if (now == k) return flip(0, lst ^ val), lst = val, ANS[val] = ans[0], void();
    solve(val, now + 1), solve(val | (1 << now), now + 1);
  }
  int main() {
    IN(n), IN(k), dep[0] = k - 1;
    for (int i = 1; i <= n; ++i) IN(a[i]), insert(0, a[i]);
    solve(0, 0);
    for (int i = 0; i < 1 << k; ++i) printf("%d ", ANS[i]);
    puts("");
    return 0;
  }
  