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;
  int j, sz[1050005], rv[1050005], mn[1050005], mx[1050005], v[1050005],
      ans[1050005], c[1050005], ot[1050005];
  inline void up(int k) {
    v[k] = v[k << 1] + v[k << 1 | 1];
    if (!v[k]) return;
    if (!v[k << 1 | 1]) {
      mn[k] = mn[k << 1];
      mx[k] = mx[k << 1];
      ans[k] = ans[k << 1];
    } else if (!v[k << 1]) {
      mn[k] = mn[k << 1 | 1] + sz[k << 1];
      mx[k] = mx[k << 1 | 1] + sz[k << 1];
      ans[k] = ans[k << 1 | 1];
    } else {
      mn[k] = mn[k << 1];
      mx[k] = mx[k << 1 | 1] + sz[k << 1];
      ans[k] = min(min(ans[k << 1], ans[k << 1 | 1]),
                   mn[k << 1 | 1] + sz[k << 1] - mx[k << 1]);
    }
  }
  void build(int k, int l, int r) {
    sz[k] = r - l + 1;
    if (l == r)
      v[k] = c[l], ans[k] = 1e9;
    else
      build(k << 1, l, (l + r >> 1)), build(k << 1 | 1, (l + r >> 1) + 1, r),
          up(k);
  }
  void update(int k, int l, int r, int x) {
    if (x == j) {
      swap(mn[k << 1], mn[k << 1 | 1]);
      swap(mx[k << 1], mx[k << 1 | 1]);
      swap(ans[k << 1], ans[k << 1 | 1]);
      swap(v[k << 1], v[k << 1 | 1]);
      up(k);
      return;
    }
    ++x;
    update(k << 1, l, (l + r >> 1), x);
    update(k << 1 | 1, (l + r >> 1) + 1, r, x);
    up(k);
  }
  int n, k, x, i;
  int main() {
    cin >> k >> n;
    for (i = 1; i <= k; ++i) cin >> x, c[x] = 1;
    build(1, 0, (1 << n) - 1);
    ot[0] = ans[1];
    for (i = 1; i < (1 << n); ++i) rv[i] = (rv[i >> 1] >> 1) | ((i & 1) << n - 1);
    for (i = 1; i < (1 << n); ot[rv[i]] = ans[1], ++i)
      for (j = 0; j < n; ++j)
        if ((rv[i] ^ rv[i - 1]) & (1 << (n - j - 1)))
          update(1, 0, (1 << n) - 1, 0);
    for (i = 0; i < (1 << n); ++i) cout << ot[i] << "\n";
  }
  