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 long long maxn = 1 << 19;
  long long n, k;
  long long ans[maxn], mx[maxn], mn[maxn];
  bool occ[maxn];
  signed main() {
    ios::sync_with_stdio(false);
    cin.tie(0);
    cin >> n >> k;
    for (long long i = 0; i < (long long)(1 << k); ++i)
      mx[i] = -1e9, mn[i] = 1e9, ans[i] = 2e9;
    for (long long i = 0; i < (long long)(n); ++i) {
      long long u;
      cin >> u;
      occ[u] = 1, mx[u] = u, mn[u] = u;
    }
    for (long long i = 0; i < k; ++i) {
      for (long long j = 0; j < (1 << k); j += (1 << i + 1)) {
        for (long long t = 0; t < (long long)(1 << i); ++t) {
          ans[j + t] = ans[j + t + (1 << i)] =
              min(ans[j + t], ans[j + t + (1 << i)]);
          ans[j + t] = min(ans[j + t], mn[j + t + (1 << i)] - mx[j + t]);
          ans[j + t + (1 << i)] =
              min(ans[j + t + (1 << i)],
                  mn[j + t] - mx[j + t + (1 << i)] + (1 << i + 1));
        }
      }
      for (long long j = 0; j < (1 << k); j += (1 << i + 1)) {
        for (long long t = 0; t < (long long)(1 << i); ++t) {
          long long upmn0 = min(mn[j + t], mn[j + t + (1 << i)]);
          long long upmn1 =
              min(mn[j + t + (1 << i)] - (1 << i), mn[j + t] + (1 << i));
          long long upmx0 = max(mx[j + t], mx[j + t + (1 << i)]);
          long long upmx1 =
              max(mx[j + t + (1 << i)] - (1 << i), mx[j + t] + (1 << i));
          mn[j + t] = upmn0, mn[j + t + (1 << i)] = upmn1;
          mx[j + t] = upmx0, mx[j + t + (1 << i)] = upmx1;
        }
      }
    }
    for (long long i = 0; i < (long long)(1 << k); ++i) cout << ans[i] << " ";
    return 0;
  }
  