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 <typename T>
  using minPq = priority_queue<T, vector<T>, greater<T>>;
  template <typename T>
  inline void UNIQUE(vector<T> &v) {
    sort((v).begin(), (v).end());
    v.resize(unique((v).begin(), (v).end()) - v.begin());
  }
  inline constexpr int lg2(int x) { return 32 - __builtin_clz(x); }
  inline constexpr long long lg2(long long x) { return 32 - __builtin_clzll(x); }
  const int INF = 0x3f3f3f3f;
  const long long LLINF = 0x3f3f3f3f3f3f3f3f;
  const double PI = acos(-1);
  int n, k;
  struct Node {
    int dif, mn, mx;
  };
  bool input[1 << 19];
  vector<Node> build(int l, int r) {
    int len = r - l;
    if (l + 1 == r) {
      vector<Node> res;
      if (input[l])
        res.push_back({INF, l, l});
      else
        res.push_back({INF, INF, INF});
      return res;
    }
    int mid = (l + r) / 2;
    vector<Node> lRes = build(l, mid);
    vector<Node> rRes = build(mid, r);
    vector<Node> ans;
    for (int i = 0; i < len / 2; ++i) {
      if (lRes[i].mn == INF && rRes[i].mn == INF)
        ans.push_back({INF, INF, INF});
      else if (lRes[i].mn == INF)
        ans.push_back(rRes[i]);
      else if (rRes[i].mn == INF)
        ans.push_back(lRes[i]);
      else {
        ans.push_back(
            {min(rRes[i].mn - lRes[i].mx, min(lRes[i].dif, rRes[i].dif)),
             lRes[i].mn, rRes[i].mx});
      }
    }
    for (int i = 0; i < len / 2; ++i) {
      if (lRes[i].mn == INF && rRes[i].mn == INF)
        ans.push_back({INF, INF, INF});
      else if (lRes[i].mn == INF)
        ans.push_back(
            {rRes[i].dif, rRes[i].mn ^ (len / 2), rRes[i].mx ^ (len / 2)});
      else if (rRes[i].mn == INF)
        ans.push_back(
            {lRes[i].dif, lRes[i].mn ^ (len / 2), lRes[i].mx ^ (len / 2)});
      else {
        ans.push_back({min((lRes[i].mn ^ (len / 2)) - (rRes[i].mx ^ (len / 2)),
                           min(lRes[i].dif, rRes[i].dif)),
                       rRes[i].mn ^ (len / 2), lRes[i].mx ^ (len / 2)});
      }
    }
    return ans;
  }
  int main() {
    cin.sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    cin >> n >> k;
    for (int i = 0; i < n; ++i) {
      int a;
      cin >> a;
      input[a] = true;
    }
    auto ans = build(0, 1 << k);
    for (int i = 0; i < (1 << k); ++i) {
      cout << ans[i].dif << " \n"[i == (1 << k) - 1];
    }
  }
  