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("Ofast")
  #pragma GCC optimize("unroll-loops")
  #pragma GCC target("avx,avx2,fma")
  #pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
  using namespace std;
  const int MAX_N = 21;
  const long long MOD = 1000000007;
  const int INF = 0x3f3f3f3f;
  int n, k;
  int lef[1 << MAX_N], rig[1 << MAX_N], mnval[1 << MAX_N], level[1 << MAX_N],
      mn[1 << MAX_N], mx[1 << MAX_N], ans[1 << MAX_N];
  int st[MAX_N], nums[1 << MAX_N];
  int main(int argc, const char* argv[]) {
    cin.tie(0)->sync_with_stdio(0);
    cin.exceptions(cin.failbit);
    cin >> n >> k;
    for (int i = (0); i < (1 << k); i++) lef[i] = 2 * i + 1, rig[i] = 2 * i + 2;
    for (int i = (0); i < (1 << (k + 1)); i++) mn[i] = INF, mx[i] = -INF;
    for (int i = (1); i < (1 << (k + 1)); i++) {
      level[i] = level[(i - 1) / 2] + 1;
      st[level[i]] = max(st[level[i]], i);
      mnval[i] = mnval[(i - 1) / 2] + ((!(i & 1)) << (k - level[i]));
    }
    for (int i = (0); i < (n); i++) {
      int x;
      cin >> x;
      int curr = 0;
      mn[0] = min(mn[0], x), mx[0] = max(mx[0], x);
      for (int j = (k)-1; j >= (0); j--) {
        if (x & (1 << j))
          curr = rig[curr];
        else
          curr = lef[curr];
        mn[curr] = min(mn[curr], x - mnval[curr]);
        mx[curr] = max(mx[curr], x - mnval[curr]);
      }
    }
    for (int i = (0); i < ((1 << (k + 2))); i++) ans[i] = INF;
    for (int i = (st[k - 1] + 1) - 1; i >= (0); i--) {
      ans[i] = min(ans[lef[i]], ans[rig[i]]);
      ans[i] =
          min(ans[i], ((mn[rig[i]] != INF && mx[lef[i]] != -INF)
                           ? (mn[rig[i]] + (1 << (k - level[i] - 1)) - mx[lef[i]])
                           : (INF)));
    }
    nums[0] = ans[0];
    for (int i = (1); i < (1 << k); i++) {
      int x = i ^ (i - 1);
      for (int j = (0); j < (k); j++)
        if (x & (1 << j)) {
          int y = st[j];
          while (~y) {
            if (level[y] == j) swap(lef[y], rig[y]);
            mn[y] = min(mn[lef[y]], mn[rig[y]] + (1 << (k - level[y] - 1)));
            mx[y] = mx[rig[y]] + (1 << (k - level[y] - 1));
            if (mx[rig[y]] == -INF) mx[y] = mx[lef[y]];
            ans[y] = min(ans[lef[y]], ans[rig[y]]);
            ans[y] =
                min(ans[y],
                    ((mn[rig[y]] != INF && mx[lef[y]] != -INF)
                         ? (mn[rig[y]] + (1 << (k - level[y] - 1)) - mx[lef[y]])
                         : (INF)));
            y--;
          }
        }
      int y = 0;
      for (int j = (0); j < (k); j++)
        if (i & (1 << j)) y += (1 << (k - j - 1));
      nums[y] = ans[0];
    }
    for (int i = (0); i < (1 << k); i++) cout << nums[i] << ' ';
    cout << endl;
    return 0;
  }
  