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>
  #pragma GCC optimize(2)
  using namespace std;
  const int maxn = (1 << 19) + 5;
  int ch[maxn << 1][2], maxi[maxn << 1], mini[maxn << 1], ans[maxn << 1];
  int k;
  int out[maxn], cnt;
  void update(int x, int l, int r) {
    int mid = (l + r) >> 1;
    maxi[x] = max(maxi[ch[x][0]], maxi[ch[x][1]] + (mid - l + 1));
    mini[x] = min(mini[ch[x][0]], mini[ch[x][1]] + (mid - l + 1));
    ans[x] = min(ans[ch[x][0]], ans[ch[x][1]]);
    ans[x] = min(ans[x], mini[ch[x][1]] + (mid - l + 1) - maxi[ch[x][0]]);
  }
  void update(int &x, int l, int r, int iter) {
    if (!x) x = ++cnt;
    if (l == r) {
      maxi[x] = mini[x] = 0;
      ans[x] = 1e9;
      return;
    }
    int mid = (l + r) >> 1;
    if (mid >= iter)
      update(ch[x][0], l, mid, iter);
    else
      update(ch[x][1], mid + 1, r, iter);
    update(x, l, r);
  }
  void swap(int x, int l, int r, int to) {
    if (!x) return;
    int mid = (l + r) >> 1;
    if ((r - l + 1) == (1 << to))
      swap(ch[x][0], ch[x][1]);
    else {
      swap(ch[x][0], l, mid, to);
      swap(ch[x][1], mid + 1, r, to);
    }
    update(x, l, r);
  }
  signed main() {
    ios::sync_with_stdio(0);
    cin.tie(0);
    int n;
    cin >> n >> k;
    mini[0] = 1e9;
    maxi[0] = -1e9;
    ans[0] = 1e9;
    int root = 0, m = (1 << k) - 1;
    for (int i = 1; i <= n; i++) {
      int x;
      cin >> x;
      update(root, 0, m, x);
    }
    cout << ans[root] << ' ';
    int lst = 0;
    for (int x = 1; x <= m; x++) {
      int to = 0;
      for (int j = 0; j < k; j++)
        if (x & (1 << j)) to ^= 1 << (k - j - 1);
      for (int j = 0; j < k; j++)
        if ((lst & (1 << j)) != (to & (1 << j))) {
          swap(root, 0, m, j + 1);
        }
      out[to] = ans[root];
      lst = to;
    }
    for (int i = 1; i <= m; i++) cout << out[i] << ' ';
    return 0;
  }
  