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;
  long long n, k;
  bool b[1 << 19];
  int32_t main() {
    cin >> n >> k;
    vector<vector<long long>> c(1 << k, vector<long long>(1, 1e9));
    vector<vector<long long>> l(1 << k, vector<long long>(1, 1e9));
    vector<vector<long long>> r(1 << k, vector<long long>(1, 1e9));
    for (long long i = 0; i < n; ++i) {
      long long a;
      cin >> a;
      l[a][0] = r[a][0] = 0;
    }
    while (c.size() > 1) {
      vector<vector<long long>> cc;
      vector<vector<long long>> ll;
      vector<vector<long long>> rr;
      for (long long i = 0; i < c.size(); i += 2) {
        vector<long long> &c1 = c[i], &c2 = c[i + 1];
        vector<long long> &l1 = l[i], &l2 = l[i + 1];
        vector<long long> &r1 = r[i], &r2 = r[i + 1];
        vector<long long> c3(2 * c1.size()), l3(2 * c1.size()), r3(2 * c1.size());
        for (long long j = 0; j < c1.size(); ++j) {
          c3[j] = c3[c1.size() + j] = min(c1[j], c2[j]);
          c3[j] = min(c3[j], r1[j] + l2[j] + 1);
          c3[c1.size() + j] = min(c3[c1.size() + j], l1[j] + r2[j] + 1);
          l3[j] = min(l1[j], (long long)c1.size() + l2[j]);
          l3[c1.size() + j] = min(l2[j], (long long)c1.size() + l1[j]);
          r3[j] = min(r2[j], (long long)c1.size() + r1[j]);
          r3[c1.size() + j] = min(r1[j], (long long)c1.size() + r2[j]);
        }
        cc.push_back(c3);
        ll.push_back(l3);
        rr.push_back(r3);
      }
      c = cc;
      l = ll;
      r = rr;
    }
    for (long long a : c[0]) cout << a << " ";
    cout << "\n";
  }
  