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 max_k = 19;
  const int max_n = (1 << max_k) + 5, inf = 1000111222;
  const int max_v = max_n * max_k;
  int n, k, a[max_n], ans[max_k][max_n];
  int first, nxt[max_v][2];
  bool get_bit(int mask, int pos) { return (mask >> pos) & 1; }
  void add(int x) {
    int v = 0;
    for (int i = k - 1; i >= 0; --i) {
      if (!nxt[v][get_bit(x, i)]) {
        nxt[v][get_bit(x, i)] = ++first;
      }
      v = nxt[v][get_bit(x, i)];
    }
  }
  int get_min(int v, int k, int x) {
    int res = 0;
    for (int i = k; i >= 0; --i) {
      int f = get_bit(x, i);
      if (!nxt[v][f]) {
        res ^= 1 << i;
        f ^= 1;
      }
      assert(nxt[v][f]);
      v = nxt[v][f];
    }
    return res;
  }
  int get_max(int v, int k, int x) {
    int res = 0;
    for (int i = k; i >= 0; --i) {
      int f = !get_bit(x, i);
      res ^= 1 << i;
      if (!nxt[v][f]) {
        res ^= 1 << i;
        f ^= 1;
      }
      assert(nxt[v][f]);
      v = nxt[v][f];
    }
    return res;
  }
  void solve(int v, int lev) {
    const int max_x = 1 << (lev + 1);
    fill(ans[lev], ans[lev] + max_x, inf);
    if (!lev) {
      if (nxt[v][0] && nxt[v][1]) {
        ans[lev][0] = 1;
        ans[lev][1] = 1;
      }
      return;
    }
    for (int i = 0; i < 2; ++i) {
      if (nxt[v][i]) {
        solve(nxt[v][i], lev - 1);
        for (int x = 0; x < max_x; ++x) {
          ans[lev][x] =
              min(ans[lev][x], ans[lev - 1][x & ((max_x - 1) ^ (1 << lev))]);
        }
      }
    }
    if (nxt[v][0] && nxt[v][1]) {
      for (int x = 0; x < max_x; ++x) {
        const int f = !get_bit(x, lev);
        ans[lev][x] =
            min(ans[lev][x], get_min(nxt[v][f], lev - 1, x) + (1 << lev) -
                                 get_max(nxt[v][!f], lev - 1, x));
      }
    }
  }
  int main() {
    scanf("%d%d", &n, &k);
    for (int i = 0; i < n; ++i) {
      scanf("%d", &a[i]);
      add(a[i]);
    }
    solve(0, k - 1);
    for (int i = 0; i < (1 << k); ++i) {
      printf("%d ", ans[k - 1][i]);
    }
    puts("");
    return 0;
  }
  