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 MAXN = 1 << 19 | 7;
  const int Inf = 1e9;
  int n, K, Top, A[MAXN];
  int F[MAXN], Minv[MAXN], Maxv[MAXN];
  int Read() {
    int a = 0, c = 1;
    char b = getchar();
    while (b != '-' && (b < '0' || b > '9')) b = getchar();
    if (b == '-') c = -1, b = getchar();
    while (b >= '0' && b <= '9') a = a * 10 + b - 48, b = getchar();
    return a * c;
  }
  int Min(int A, int B) { return A < B ? A : B; }
  int Max(int A, int B) { return A > B ? A : B; }
  int main() {
    n = Read(), K = Read(), Top = (1 << K) - 1;
    for (int i = 0; i <= Top; i++) F[i] = Minv[i] = Inf, Maxv[i] = -Inf;
    for (int i = 1; i <= n; i++) A[i] = Read(), Minv[A[i]] = Maxv[A[i]] = 0;
    for (int Nb = 0, Bit = 1; Nb < K; Nb++, Bit <<= 1)
      for (int Ri = 0, Le; Ri <= Top; Ri++) {
        if (~Ri & Bit) continue;
        Le = Ri ^ Bit, F[Le] = F[Ri] = Min(F[Le], F[Ri]);
        int Minr = Minv[Ri], Maxr = Maxv[Ri], Minl = Minv[Le], Maxl = Maxv[Le];
        F[Le] = Min(F[Le], Minr + Bit - Maxl),
        F[Ri] = Min(F[Ri], Minl + Bit - Maxr);
        Minv[Le] = Min(Minl, Minr + Bit), Maxv[Le] = Max(Maxl, Maxr + Bit);
        Minv[Ri] = Min(Minr, Minl + Bit), Maxv[Ri] = Max(Maxr, Maxl + Bit);
      }
    for (int i = 0; i <= Top; i++) printf("%d ", F[i]);
  }
  