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;
  void rec(vector<int>& mi, int base, int opt, const vector<int>& a, int al,
           int ar, int p) {
    if (p == -1) {
      assert(ar - al == 1);
      int i = opt + 1;
      do {
        i = (i - 1) & opt;
        int x = base + i;
        mi[x] = a[al] ^ x;
      } while (i);
      return;
    }
    int am = ar;
    for (int i = al; i < ar; i++) {
      if (a[i] >> p & 1) {
        am = i;
        break;
      }
    }
    if (am == al) {
      rec(mi, base, opt, a, al, ar, p - 1);
    } else if (am == ar) {
      rec(mi, base, opt, a, al, ar, p - 1);
    } else {
      rec(mi, base, opt ^ (1 << p), a, al, am, p - 1);
      rec(mi, base | 1 << p, opt ^ (1 << p), a, am, ar, p - 1);
    }
  }
  vector<int> recmi(const vector<int>& a, int p) {
    vector<int> mi(1 << (p + 1));
    rec(mi, 0, (1 << (p + 1)) - 1, a, 0, a.size(), p);
    return mi;
  }
  vector<int> recma(vector<int> a, int p) {
    for (int& v : a) v ^= (1 << (p + 1)) - 1;
    reverse(a.begin(), a.end());
    vector<int> mi = recmi(a, p);
    for (int& v : mi) v ^= (1 << (p + 1)) - 1;
    return mi;
  }
  vector<int> slowrecma(vector<int> a, int p) {
    vector<int> ma(1 << (p + 1), -1);
    for (int x = 0; x < 1 << (p + 1); x++)
      for (int v : a) ma[x] = max(ma[x], v ^ x);
    return ma;
  }
  vector<int> slowrecmi(vector<int> a, int p) {
    vector<int> mi(1 << (p + 1), 1e8);
    for (int x = 0; x < 1 << (p + 1); x++)
      for (int v : a) mi[x] = min(mi[x], v ^ x);
    return mi;
  }
  int ans[20][1 << 19];
  void rec2(const vector<int>& a, int p) {
    if (a.size() <= 1) return;
    assert(p >= 0);
    vector<int> l, r;
    for (int v : a) {
      (v >> p & 1 ? r : l).push_back(v);
    }
    if (l.size() && r.size()) {
      auto lmi = recmi(l, p - 1);
      auto lma = recma(l, p - 1);
      auto rmi = recmi(r, p - 1);
      auto rma = recma(r, p - 1);
      for (int x = 0; x < 1 << (p + 1); x++) {
        if (x >> p & 1) {
          ans[p][x] = min(ans[p][x], ((lmi[x - (1 << p)] + (1 << p)) -
                                      (rma[x - (1 << p)] - (1 << p))));
        } else {
          ans[p][x] = min(ans[p][x], (rmi[x] - lma[x]));
        }
      }
    }
    rec2(l, p - 1);
    rec2(r, p - 1);
  }
  int main() {
    ios::sync_with_stdio(0);
    cin.tie(0);
    int n, k;
    cin >> n >> k;
    vector<int> a(n);
    for (int& v : a) cin >> v;
    sort(a.begin(), a.end());
    for (int i = 0; i < k; i++)
      for (int j = 0; j < 1 << k; j++) ans[i][j] = 1e9;
    rec2(a, k - 1);
    for (int x = 0; x < 1 << k; x++) {
      int mi = 1e9;
      for (int i = 0; i < k; i++) mi = min(mi, ans[i][x & ((1 << (i + 1)) - 1)]);
      cout << mi << ' ';
    }
    cout << endl;
  }
  