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>
  template <typename T>
  bool ckmax(T &x, T y) {
    return x < y ? x = y, true : false;
  }
  template <typename T>
  bool ckmin(T &x, T y) {
    return x > y ? x = y, true : false;
  }
  using namespace std;
  const long long inf = 1000000000;
  const double Pi = acos(-1);
  const long long mod = 998244353;
  const double eps = 1e-6;
  inline long long fpow(long long a, long long b = mod - 2, long long p = mod) {
    a %= p;
    long long res = 1;
    while (b) {
      if (b & 1) res = 1ll * res * a % p;
      a = 1ll * a * a % p;
      b >>= 1;
    }
    return res;
  }
  inline long long read() {
    char c = getchar();
    long long x = 0;
    bool f = 0;
    for (; !isdigit(c); c = getchar()) f ^= !(c ^ 45);
    for (; isdigit(c); c = getchar()) x = (x << 1) + (x << 3) + (c ^ 48);
    if (f) x = -x;
    return x;
  }
  inline void write(long long x) {
    if (x < 0) {
      putchar('-');
      write(-x);
      return;
    }
    if (x >= 10) write(x / 10);
    putchar(x % 10 + '0');
  }
  inline void writesp(long long x) { write(x), putchar(' '); }
  inline void writeln(long long x) {
    write(x);
    putchar('\n');
  }
  long long k, n;
  long long ans[2222222], mn[2][2222222], mx[2][2222222];
  signed main() {
    n = read(), k = read();
    for (long long i = (0), iE = ((1 << k) + 5); i <= iE; i++)
      mn[1][i] = ans[i] = inf, mx[1][i] = -inf;
    for (long long i = (1), iE = (n); i <= iE; i++) {
      static long long x;
      x = read();
      mn[1][x] = mx[1][x] = 0;
    }
    for (long long i = (0), iE = (k - 1); i <= iE; i++)
      for (long long j = (0), jE = ((1 << k) - 1); j <= jE; j++)
        if (j & (1 << i)) {
          long long k = j ^ (1 << i);
          ans[j] = ans[k] = min(ans[j], ans[k]);
          ckmin(ans[j], mn[(i + 1) & 1][k] - mx[(i + 1) & 1][j] + (1ll << i));
          ckmin(ans[k], mn[(i + 1) & 1][j] - mx[(i + 1) & 1][k] + (1ll << i));
          mn[i & 1][j] = min(mn[(i + 1) & 1][j], mn[(i + 1) & 1][k] + (1ll << i));
          mn[i & 1][k] = min(mn[(i + 1) & 1][k], mn[(i + 1) & 1][j] + (1ll << i));
          mx[i & 1][j] = max(mx[(i + 1) & 1][j], mx[(i + 1) & 1][k] + (1ll << i));
          mx[i & 1][k] = max(mx[(i + 1) & 1][k], mx[(i + 1) & 1][j] + (1ll << i));
        }
    for (long long i = (0), iE = ((1 << k) - 1); i <= iE; i++) writesp(ans[i]);
  }
  