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 K = 19, N = 1 << K, INF = 0x3f3f3f3f;
  int n, k, cnt[N], a[N], ans[K][N];
  vector<int> b[N], c[N];
  void ckmin(int &x, int y) {
    if (y < x) x = y;
  }
  void calc(vector<int> v, int s, int t) {
    if (v.size() < 2) return;
    if (s < t) {
      for (int i = 0; i < 1 << t; i++)
        ckmin(ans[s][i], abs((v[0] ^ i) - (v[1] ^ i)));
      return;
    }
    for (int i = 0; i < 1 << (s + 1); i++) c[i].clear();
    for (int i = 0; i < 1 << t; i++)
      for (auto u : v) {
        int to = i ^ (1 << s);
        c[to ^ u].push_back(to), to = i ^ (((1 << (s - t)) - 1) << t),
                                 c[to ^ u].push_back(to);
      }
    for (int i = 0; i < 1 << (s + 1); i++) {
      int mn = INF;
      sort(c[i].begin(), c[i].end());
      for (int j = 1; j < c[i].size(); j++) ckmin(mn, c[i][j] - c[i][j - 1]);
      ckmin(ans[s][i], mn);
    }
    return;
  }
  int main() {
    scanf("%d%d", &n, &k);
    for (int i = 0; i < n; i++) scanf("%d", &a[i]);
    int p = 0;
    for (int i = 1; i <= k; i++) {
      int mx = 0;
      for (int j = 0; j < 1 << k; j++) cnt[j] = 0;
      for (int j = 0; j < n; j++) cnt[a[j] >> i]++;
      for (int j = 0; j < 1 << k; j++) mx = max(mx, cnt[j]);
      if (mx > 1) {
        p = i;
        break;
      }
    }
    memset(ans, 0x3f, sizeof(ans));
    for (int i = p - 1; i < k; i++) {
      for (int j = 0; j < 1 << (k - i - 1); j++) b[j].clear();
      for (int j = 0; j < n; j++)
        b[a[j] >> (i + 1)].push_back(a[j] & ((1 << (i + 1)) - 1));
      for (int j = 0; j < 1 << (k - i - 1); j++) calc(b[j], i, p);
    }
    for (int i = 0; i < 1 << k; i++) {
      int mn = INF;
      for (int j = p - 1; j < k; j++) ckmin(mn, ans[j][i & ((1 << (j + 1)) - 1)]);
      printf("%d ", mn);
    }
  }
  