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;
  using ll = long long;
  using ull = long unsigned long;
  using ld = double long;
  const int inf = 1e9;
  struct the_tree {
    struct node {
      struct kec {};
      struct nula {};
      int l, r, s, g;
      node() : l(-1), r(-1), s(0), g(1e9) {}
      node(kec) : l(0), r(0), s(1), g(1e9) {}
      node(nula) : l(-1), r(-1), s(1), g(1e9) {}
      node(int l, int r, int s, int g) : l(l), r(r), s(s), g(g) {}
      node operator+(const node& b) const {
        if (l == -1 && b.l == -1) {
          return node{-1, -1, s + b.s, inf};
        } else if (l == -1) {
          return {s + b.l, s + b.r, s + b.s, b.g};
        } else if (b.l == -1) {
          return {l, r, s + b.s, g};
        } else {
          return {l, s + b.r, s + b.s, min({g, b.g, s + b.l - r})};
        }
      }
    };
    vector<node> a;
    vector<int> fl;
    int k, n;
    the_tree(int k, const vector<int>& b) {
      this->k = k;
      n = 1 << k;
      a.resize(2 * n, node::nula());
      fl.resize(k);
      for (int x : b) {
        a[x + n] = node::kec();
      }
      for (int i = n - 1; i; i--) a[i] = a[2 * i] + a[2 * i + 1];
    }
    int operator()() const { return a[1].g; }
    void fetch(int x) {
      if (x >= n) return;
      int lvl = k - 1 - (31 - __builtin_clz(x));
      if (fl[lvl]) {
        a[x] = a[2 * x + 1] + a[2 * x];
      } else {
        a[x] = a[2 * x] + a[2 * x + 1];
      }
    }
    void flip(int b) {
      fl[b] ^= 1;
      for (int x = (1 << (k - b)) - 1; x; x--) {
        fetch(x);
      }
    }
  };
  int n, k;
  int ans[1 << 19];
  int main() {
    ios::sync_with_stdio(!cin.tie(0));
    cin >> n >> k;
    vector<int> a(n);
    for (int i = 0; i < n; i++) cin >> a[i];
    the_tree drvo(k, a);
    ans[0] = drvo();
    for (int i = 1, mask = 0; i < (1 << k); i++) {
      int b = k - 1 - __builtin_ctz(i);
      mask ^= 1 << b;
      drvo.flip(b);
      ans[mask] = drvo();
    }
    for (int i = 0; i < (1 << k); i++) {
      cout << ans[i] << " \n"[i == (1 << k) - 1];
    }
  }
  