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;
  const int inf = (int)1.01e9;
  const long long infll = (long long)1.01e18;
  const long double eps = 1e-9;
  const long double pi = acos((long double)-1);
  mt19937 mrand(chrono::steady_clock::now().time_since_epoch().count());
  int rnd(int x) { return mrand() % x; }
  void precalc() {}
  const int maxk = 20, powk = (1 << maxk);
  int n, k;
  int x[powk];
  bool read() {
    if (scanf("%d%d", &n, &k) < 2) {
      return false;
    }
    for (int i = 0; i < n; ++i) {
      scanf("%d", &x[i]);
    }
    return true;
  }
  const int maxvs = powk * (maxk + 5);
  int go[maxvs][2];
  int vs;
  int a[powk], b[powk], c[powk];
  int ans[powk];
  int newNode() {
    go[vs][0] = -1;
    go[vs][1] = -1;
    return vs++;
  }
  void add(int x, int k) {
    int v = 0;
    for (int i = k - 1; i >= 0; --i) {
      int d = ((x >> i) & 1);
      if (go[v][d] == -1) {
        go[v][d] = newNode();
      }
      v = go[v][d];
    }
  }
  int getMin(int v, int x, int y, int i) {
    if (i < 0) {
      return x;
    }
    int d = ((y >> i) & 1);
    if (go[v][d] != -1) {
      return getMin(go[v][d], x, y, i - 1);
    }
    return getMin(go[v][!d], x ^ (1 << i), y, i - 1);
  }
  int getMax(int v, int x, int y, int i) {
    if (i < 0) {
      return x;
    }
    int d = ((y >> i) & 1);
    if (go[v][!d] != -1) {
      return getMax(go[v][!d], x ^ (1 << i), y, i - 1);
    }
    return getMax(go[v][d], x, y, i - 1);
  }
  void solve1(int k) {
    vs = 0;
    newNode();
    for (int i = 0; i < (1 << k); ++i) {
      if (b[i]) {
        add(i, k);
      }
    }
    if (go[0][0] == -1 || go[0][1] == -1) {
      return;
    }
    for (int i = 0; i < (1 << k); ++i) {
      int v = go[0][0], u = go[0][1];
      if (i & (1 << (k - 1))) {
        swap(v, u);
      }
      int res = getMin(u, (1 << (k - 1)), i, k - 2) - getMax(v, 0, i, k - 2);
      c[i] = min(c[i], res);
    }
  }
  void solve() {
    for (int i = 0; i < (1 << k); ++i) {
      a[i] = false;
      ans[i] = inf;
    }
    for (int i = 0; i < n; ++i) {
      a[x[i]] = true;
    }
    for (int l = 0; l < k; ++l) {
      for (int i = 0; i < (1 << (k - l)); ++i) {
        c[i] = inf;
      }
      for (int i = 0; i < (1 << l); ++i) {
        for (int j = 0; j < (1 << (k - l)); ++j) {
          b[j] = a[(i << (k - l)) ^ j];
        }
        solve1(k - l);
      }
      for (int i = 0; i < (1 << l); ++i) {
        for (int j = 0; j < (1 << (k - l)); ++j) {
          auto &cur = ans[(i << (k - l)) ^ j];
          cur = min(cur, c[j]);
        }
      }
    }
    for (int i = 0; i < (1 << k); ++i) {
      printf("%d ", ans[i]);
    }
    printf("\n");
  }
  int main() {
    precalc();
    while (read()) {
      solve();
    }
    return 0;
  }
  