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 + 10;
  const long long mod = 1e9 + 7;
  inline long long gcd(long long a, long long b) {
    return !b ? a : gcd(b, a % b);
  }
  inline long long q_pow(long long a, long long x) {
    long long ans = 1, tmp = a;
    while (x) {
      if (x & 1) (ans *= tmp) %= mod;
      (tmp *= tmp) %= mod;
      x >>= 1;
    }
    return ans;
  }
  template <typename T>
  inline void re(T &N) {
    int f = 1;
    char c;
    while ((c = getchar()) < '0' || c > '9')
      if (c == '-') f = -1;
    N = c - '0';
    while ((c = getchar()) >= '0' && c <= '9') N = N * 10 + c - '0';
    N *= f;
  }
  int m, n, t = 1, st, en;
  struct Trie {
    struct node {
      int a[2];
      int ans, len;
      int ma, mi;
      node() {
        a[0] = a[1] = 0;
        ans = 1e9;
        len = 1;
        ma = -1e9;
        mi = 1e9;
      }
      node(int x) {
        a[0] = a[1] = 0;
        ans = 1e9;
        len = 1;
        ma = mi = 0;
      }
    } nul;
    vector<node> v;
    Trie() { v.assign(1, nul); }
    inline void push_up(int p) {
      v[p].mi = min(v[(v[p].a[0])].mi, v[(v[p].a[1])].mi + v[(v[p].a[0])].len);
      v[p].ma = max(v[(v[p].a[0])].ma, v[(v[p].a[1])].ma + v[(v[p].a[0])].len);
      v[p].ans =
          min({v[(v[p].a[0])].ans, v[(v[p].a[1])].ans,
               v[(v[p].a[1])].mi - v[(v[p].a[0])].ma + v[(v[p].a[0])].len});
      v[p].len = v[(v[p].a[0])].len + v[(v[p].a[1])].len;
    }
    int build(int p, int l, int r) {
      if (!p) v.push_back(nul), p = (int)v.size() - 1;
      if (l == r) return p;
      int mid = (l + r) >> 1;
      (v[p].a[0]) = build((v[p].a[0]), l, mid);
      (v[p].a[1]) = build((v[p].a[1]), mid + 1, r);
      push_up(p);
      return p;
    }
    void insert(int p, int dep, int val) {
      if (!(~dep)) return v[p] = node(1), void();
      insert(v[p].a[val >> dep & 1], dep - 1, val);
      push_up(p);
    }
    void rotate(int p, int dep, int lim) {
      if (dep == lim) {
        swap((v[p].a[0]), (v[p].a[1]));
        push_up(p);
        return;
      }
      rotate((v[p].a[0]), dep - 1, lim);
      rotate((v[p].a[1]), dep - 1, lim);
      push_up(p);
    }
  } tr;
  int main() {
    ios::sync_with_stdio(0);
    int n, k;
    cin >> n >> k;
    vector<int> a(n);
    for (int i = 0; i < n; i++) cin >> a[i];
    tr.build(0, 0, (1 << k) - 1);
    for (int i = 0; i < n; i++) tr.insert(1, k - 1, a[i]);
    vector<int> ans(1 << k), gray(1 << k);
    for (int i = 0; i < 1 << k; i++) {
      gray[i] = (i >> 1) ^ i;
      vector<int> tmp;
      for (int j = 0; j < k; j++, gray[i] >>= 1) tmp.push_back(gray[i] & 1);
      for (int j = 0; j < k; j++) gray[i] = gray[i] << 1 | tmp[j];
    }
    for (int i = 0; i < 1 << k; i++) {
      ans[gray[i]] = tr.v[1].ans;
      if ((i + 1) < 1 << k) {
        int pos = 31 - __builtin_clz(gray[i] ^ gray[i + 1]);
        tr.rotate(1, k - 1, pos);
      }
    }
    for (int i = 0; i < 1 << k; i++) cout << ans[i] << " \n"[i + 1 == 1 << k];
  }
  