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;
  template <class T>
  vector<T> readvec(int N) {
    vector<T> res(N);
    for (int i = 0; i < N; ++i) cin >> res[i];
    return res;
  }
  void one() {
    int N, K;
    cin >> N >> K;
    vector<int> t = readvec<int>(N);
    vector<int> mins(1 << K, 1 << K);
    vector<vector<bool>> exists(K + 1);
    for (int i = 0; i <= K; ++i) {
      exists[i].resize(1 << i, false);
    }
    for (int j : t) exists[K][j] = true;
    for (int k = K - 1; k >= 0; --k) {
      for (int j = 0; j < (1 << k); ++j) {
        exists[k][j] = exists[k + 1][j * 2] | exists[k + 1][j * 2 + 1];
      }
    }
    for (int k = 0; k < K; ++k) {
      for (int i = 0; i < (1 << (K - k - 1)); ++i) {
        int val = 1 << K;
        for (int j = 0; j < (1 << (k + 1)); j += 2) {
          if (exists[k + 1][j] && exists[k + 1][j + 1]) {
            int d = 1;
            int l = j;
            int r = j + 1;
            for (int cb = k + 1; cb < K; ++cb) {
              int currb = ((i >> (K - cb - 1)) & 1);
              d *= 2;
              l *= 2;
              r *= 2;
              if (exists[cb + 1][l ^ 1 ^ currb] && exists[cb + 1][r ^ currb]) {
                r ^= currb;
                l ^= currb ^ 1;
                d -= 1;
              } else if (exists[cb + 1][l] && exists[cb + 1][r]) {
              } else if (exists[cb + 1][l ^ 1] && exists[cb + 1][r ^ 1]) {
                l ^= 1;
                r ^= 1;
              } else {
                r ^= currb ^ 1;
                l ^= currb;
                d += 1;
              }
            }
            assert(d == (r ^ i) - (l ^ i));
            val = min(val, d);
          }
        }
        for (int j = 0; j < (1 << k); ++j) {
          mins[(j << (K - k)) + i] = min(mins[(j << (K - k)) + i], val);
          mins[(j << (K - k)) + (1 << (K - k)) - 1 - i] =
              min(mins[(j << (K - k)) + (1 << (K - k)) - 1 - i], val);
        }
      }
    }
    for (int i = 0; i < (1 << K); ++i) {
      cout << mins[i] << " \n"[i + 1 == (1 << K)];
    }
  }
  int main() {
    std::ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    one();
    cout << flush;
  }
  