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;
  using LL = long long;
  const int MAXN = 1100000;
  int l[MAXN], r[MAXN];
  int f[MAXN], ld[MAXN], rd[MAXN];
  void Build(int t, int depth) {
    f[t] = ld[t] = rd[t] = MAXN;
    if (depth == 0) return;
    l[t] = t + t;
    r[t] = t + t + 1;
    Build(t + t, depth - 1);
    Build(t + t + 1, depth - 1);
  }
  void Update(int t, int B) {
    f[t] = min(f[l[t]], f[r[t]]);
    f[t] = min(f[t], rd[l[t]] + ld[r[t]] + 1);
    if (ld[l[t]] == MAXN) {
      ld[t] = min(MAXN, ld[r[t]] + (1 << B));
    } else {
      ld[t] = ld[l[t]];
    }
    if (rd[r[t]] == MAXN) {
      rd[t] = min(MAXN, rd[l[t]] + (1 << B));
    } else {
      rd[t] = rd[r[t]];
    }
  }
  void Add(int t, int B, int a) {
    if (B == -1) {
      ld[t] = rd[t] = 0;
      return;
    }
    if (a >> B & 1) {
      Add(r[t], B - 1, a);
    } else {
      Add(l[t], B - 1, a);
    }
    Update(t, B);
  }
  void Change(int t, int B, int depth) {
    if (depth == 0) {
      swap(l[t], r[t]);
      Update(t, B);
      return;
    }
    Change(l[t], B - 1, depth - 1);
    Change(r[t], B - 1, depth - 1);
    Update(t, B);
  }
  int main() {
    ios::sync_with_stdio(false);
    cin.tie(0);
    int n, k;
    cin >> n >> k;
    Build(1, k);
    for (int i = 0; i < n; ++i) {
      int a;
      cin >> a;
      Add(1, k - 1, a);
    }
    if (k == 1) {
      cout << 1 << " " << 1 << endl;
      return 0;
    }
    vector<int> mask{0, 2, 3, 1};
    for (int i = 3; i <= k; ++i) {
      for (int j = (1 << (i - 1)) - 1; j >= 0; --j) {
        mask.push_back(mask[j]);
      }
      for (int j = 0; j < (1 << (i - 1)); ++j) {
        mask[j] = mask[j] << 1;
      }
      for (int j = (1 << (i - 1)); j < (1 << i); ++j) {
        mask[j] = (mask[j] << 1) | 1;
      }
    }
    vector<int> ans(1 << k);
    ans[0] = f[1];
    for (int i = 1; i < (1 << k); ++i) {
      int t = 0;
      while (((mask[i] ^ mask[i - 1]) >> t & 1) == 0) ++t;
      Change(1, k - 1, k - 1 - t);
      ans[mask[i]] = f[1];
    }
    for (int i : ans) cout << i << " ";
    cout << endl;
    return 0;
  }
  