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;
  const int N = 1e6 + 1e3 + 7, INF = 1e9;
  int cnt;
  struct T {
    int l, r;
    int ls, rs;
    int ld, rd;
    int dis;
    int sz;
  } t[N * 2 + 1];
  void update(int x) {
    t[x].ld = min({t[t[x].ls].ld, t[t[x].rs].ld + t[t[x].ls].sz, INF});
    t[x].rd = min({t[t[x].rs].rd, t[t[x].ls].rd + t[t[x].rs].sz, INF});
    t[x].sz = t[t[x].ls].sz + t[t[x].rs].sz;
    t[x].dis =
        min({t[t[x].ls].dis, t[t[x].rs].dis, t[t[x].rs].ld + t[t[x].ls].rd});
  }
  int build(int l, int r) {
    int x = ++cnt;
    t[x].l = l, t[x].r = r;
    if (l == r) {
      t[x].sz = 1;
      t[x].ld = t[x].rd = 1e9;
      t[x].dis = 1e9;
      return x;
    }
    int mid = (l + r) >> 1;
    t[x].ls = build(l, mid);
    t[x].rs = build(mid + 1, r);
    update(x);
    return x;
  }
  void ins(int x, int v) {
    if (t[x].sz == 1) {
      t[x].ld = t[x].rd = 1;
      return;
    }
    int mid = (t[x].l + t[x].r) >> 1;
    if (v <= mid)
      ins(t[x].ls, v);
    else
      ins(t[x].rs, v);
    update(x);
  }
  void change(int x, int b, int v) {
    if ((1 << b) & v) {
      swap(t[x].ls, t[x].rs);
      update(x);
      return;
    }
    change(t[x].ls, b - 1, v);
    change(t[x].rs, b - 1, v);
    update(x);
  }
  vector<int> v;
  int ans[N], n, k;
  void gen() {
    v.push_back(0);
    for (int i = 1; i <= k; i++) {
      for (auto &x : v) x <<= 1;
      int sz = v.size();
      for (int j = sz - 1; j >= 0; j--) v.push_back(v[j] ^ 1);
    }
  }
  map<int, int> bts;
  int main() {
    scanf("%d%d", &n, &k);
    gen();
    build(0, (1 << k) - 1);
    for (int i = 1; i <= n; i++) {
      int x;
      scanf("%d", &x);
      ins(1, x);
    }
    ans[0] = t[1].dis;
    for (int i = 1; i < v.size(); i++) {
      change(1, k - 1, v[i] ^ v[i - 1]), ans[v[i]] = t[1].dis;
    }
    for (int i = 0; i < (1 << k); i++)
      printf("%d%c", ans[i] - 1, " \n"[i + 1 == 1 << k]);
  }
  