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;
  vector<array<int, 3>> f(int k, vector<int> &a, int l, int r, int s) {
    if (k == 0) {
      return vector<array<int, 3>>(1);
    }
    int kk = 1 << (k - 1);
    int m;
    for (m = l; m < r; ++m)
      if (a[m] >= s + kk) break;
    vector<array<int, 3>> res(2 * kk);
    vector<array<int, 3>> u = f(k - 1, a, l, m, s);
    vector<array<int, 3>> v = f(k - 1, a, m, r, s + kk);
    for (int i = 0; i < kk; ++i) {
      int x = int(1e9);
      if (m - l > 1) x = min(x, u[i][0]);
      if (r - m > 1) x = min(x, v[i][0]);
      if (m - l > 0 && r - m > 0) x = min(x, v[i][1] + kk - u[i][2]);
      int y = int(1e9);
      int z = -1;
      if (m - l > 0) {
        y = min(y, u[i][1]);
        z = max(z, u[i][2]);
      }
      if (r - m > 0) {
        y = min(y, v[i][1] + kk);
        z = max(z, v[i][2] + kk);
      }
      res[i][0] = x;
      res[i][1] = y;
      res[i][2] = z;
    }
    for (int i = 0; i < kk; ++i) {
      int x = int(1e9);
      if (m - l > 1) x = min(x, u[i][0]);
      if (r - m > 1) x = min(x, v[i][0]);
      if (m - l > 0 && r - m > 0) x = min(x, u[i][1] + kk - v[i][2]);
      int y = int(1e9);
      int z = -1;
      if (m - l > 0) {
        y = min(y, u[i][1] + kk);
        z = max(z, u[i][2] + kk);
      }
      if (r - m > 0) {
        y = min(y, v[i][1]);
        z = max(z, v[i][2]);
      }
      res[i + kk][0] = x;
      res[i + kk][1] = y;
      res[i + kk][2] = z;
    }
    return res;
  }
  void solve() {
    int n, k;
    scanf("%d%d", &n, &k);
    vector<int> a(n);
    for (int i = 0; i < n; ++i) scanf("%d", &a[i]);
    sort(a.begin(), a.end());
    vector<array<int, 3>> ans = f(k, a, 0, a.size(), 0);
    for (int i = 0; i < (1 << k); ++i) printf("%d ", ans[i][0]);
    printf("\n");
  }
  int main() {
    solve();
    return 0;
  }
  