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;
  std::mt19937 rnd(
      (int)std::chrono::steady_clock::now().time_since_epoch().count());
  long long gcd(long long a, long long b) { return b == 0 ? a : gcd(b, a % b); }
  const int MAXBITS = 19;
  const int MAXN = 1 << MAXBITS;
  struct Res {
    int mn, mx, best;
    Res() {}
    Res(int mn, int mx, int best) : mn(mn), mx(mx), best(best) {}
  };
  int n, nbits;
  int a[MAXN];
  int ans[1 << MAXBITS];
  int cnt[1 << MAXBITS];
  Res combine(const Res& a, const Res& b, int step) {
    Res ret;
    ret.mn = a.mn != INT_MAX ? a.mn : b.mn != INT_MAX ? step + b.mn : INT_MAX;
    ret.mx = b.mx != INT_MIN ? step + b.mx : a.mx != INT_MIN ? a.mx : INT_MIN;
    ret.best = min(a.best, b.best);
    if (a.mx != INT_MIN && b.mn != INT_MAX)
      ret.best = min(ret.best, step + b.mn - a.mx);
    return ret;
  }
  vector<Res> rec(int offset, int k) {
    vector<Res> ret(1 << k);
    if (k == 0) {
      if (cnt[offset] == 0)
        ret[0] = Res(INT_MAX, INT_MIN, INT_MAX);
      else
        ret[0] = Res(0, 0, INT_MAX);
    } else {
      vector<Res> l = rec(offset, k - 1);
      vector<Res> r = rec(offset + (1 << (k - 1)), k - 1);
      for (int i = (0); i < (1 << (k - 1)); ++i) {
        ret[i] = combine(l[i], r[i], 1 << (k - 1));
        ret[(1 << (k - 1)) + i] = combine(r[i], l[i], 1 << (k - 1));
      }
    }
    return ret;
  }
  void solve() {
    for (int i = (0); i < (1 << nbits); ++i) cnt[i] = 0;
    for (int i = (0); i < (n); ++i) ++cnt[a[i]];
    for (int i = (0); i < (1 << nbits); ++i)
      if (cnt[i] >= 2) {
        for (int j = (0); j < (1 << nbits); ++j) ans[i] = 0;
        return;
      }
    vector<Res> res = rec(0, nbits);
    for (int i = (0); i < (1 << nbits); ++i) ans[i] = res[i].best;
  }
  void run() {
    scanf("%d%d", &n, &nbits);
    for (int i = (0); i < (n); ++i) scanf("%d", &a[i]);
    solve();
    for (int i = (0); i < (1 << nbits); ++i) {
      if (i != 0) printf(" ");
      printf("%d", ans[i]);
    }
    puts("");
  }
  int main() {
    run();
    return 0;
  }
  