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 std::min;
  using std::swap;
  int s[11111111][2], dl[11111111], dr[11111111], minv[11111111];
  int sz[11111111];
  int k, n;
  int rv[1 << 21];
  int a[1 << 21];
  int cp = 1;
  void pushu(int p) {
    if (s[p][0] && s[p][1])
      dl[p] = dl[s[p][0]], dr[p] = dr[s[p][1]],
      minv[p] =
          min(dr[s[p][0]] + dl[s[p][1]] + 1, min(minv[s[p][0]], minv[s[p][1]]));
    else if (s[p][0])
      dl[p] = dl[s[p][0]], dr[p] = dr[s[p][0]] + sz[p], minv[p] = minv[s[p][0]];
    else if (s[p][1])
      dr[p] = dr[s[p][1]], dl[p] = dl[s[p][1]] + sz[p], minv[p] = minv[s[p][1]];
  }
  void ins(int x, int p = 1, int i = k - 1) {
    if (i < 0) {
      dl[p] = dr[p] = 0, minv[p] = 1e9;
      return;
    }
    sz[p] = 1 << i;
    int d = (x >> i) & 1;
    if (!s[p][d]) s[p][d] = ++cp;
    ins(x - (d << i), s[p][d], i - 1), pushu(p);
  }
  void init() {
    for (register int i = 1; i <= n; i++) ins(a[i]);
  }
  void flip(int dep, int p = 1, int i = k - 1) {
    if (!p) return;
    if (i == dep) {
      swap(s[p][0], s[p][1]), pushu(p);
      return;
    }
    flip(dep, s[p][0], i - 1), flip(dep, s[p][1], i - 1), pushu(p);
  }
  int T;
  int recalc(int S) {
    for (register int i = 0; i < k; i++)
      if (((S ^ T) >> i) & 1) flip(i);
    T = S;
    return minv[1];
  }
  int ans[1 << 21];
  int main() {
    scanf("%d%d", &n, &k);
    for (register int i = 1; i <= n; i++) scanf("%d", a + i);
    init();
    for (register int i = 1; i < (1 << k); i++)
      rv[i] = (rv[i >> 1] >> 1) | ((i & 1) ? (1 << (k - 1)) : 0);
    for (register int i = 0; i < (1 << k); i++) ans[rv[i]] = recalc(rv[i]);
    for (register int i = 0; i < (1 << k); i++) printf("%d ", ans[i]);
    puts("");
  }
  