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;
  template <typename T>
  void read(T &x) {
    x = 0;
    char c = getchar();
    int f = 0;
    for (; !isdigit(c); c = getchar()) f |= c == '-';
    for (; isdigit(c); c = getchar()) x = x * 10 + (c ^ '0');
    if (f) x = -x;
  }
  template <typename T>
  void write(T x, char ed = '\n') {
    if (x < 0) putchar('-'), x = -x;
    static short st[30], tp;
    do st[++tp] = x % 10, x /= 10;
    while (x);
    while (tp) putchar(st[tp--] | '0');
    putchar(ed);
  }
  const int N = 2000500;
  int ch[N][2], f[N], a[N], lim, cnt, n, k;
  void dfs(int p, int dep, int l, int r) {
    if ((!p && dep < k) || l == r) return;
    int mid = (l + r) >> 1;
    dfs(ch[p][0], dep - 1, l, mid);
    dfs(ch[p][1], dep - 1, mid + 1, r);
    int len = 1 << (dep - 1), *f = ::f + l;
    if (!ch[p][0] || !ch[p][1]) {
      for (int i = 0; i < len; ++i) f[i] = f[i + len] = min(f[i], f[i + len]);
      return;
    }
    for (int i = 0; i < len; ++i) {
      int lmx = 0, lmn = 0, rmx = 0, rmn = 0;
      int l1 = ch[p][0], l2 = l1;
      int r1 = ch[p][1], r2 = r1;
      for (int j = dep - 2; j >= 0; --j) {
        int c = i >> j & 1, bt = 1 << j;
        if (ch[l1][c])
          l1 = ch[l1][c];
        else
          l1 = ch[l1][!c], lmn += bt;
        if (ch[l2][!c])
          l2 = ch[l2][!c], lmx += bt;
        else
          l2 = ch[l2][c];
        if (ch[r1][c])
          r1 = ch[r1][c];
        else
          r1 = ch[r1][!c], rmn += bt;
        if (ch[r2][!c])
          r2 = ch[r2][!c], rmx += bt;
        else
          r2 = ch[r2][c];
      }
      f[i] = min(f[i], f[i + len]);
      f[i + len] = min(f[i], len + lmn - rmx);
      f[i] = min(f[i], len + rmn - lmx);
    }
  }
  int main() {
    read(n), read(k), lim = 1 << k;
    for (int i = 1, x; i <= n; ++i) {
      read(x);
      int p = 0;
      for (int j = k - 1; j >= 0; --j) {
        int c = x >> j & 1;
        if (!ch[p][c]) ch[p][c] = ++cnt;
        p = ch[p][c];
      }
    }
    memset(f, 0x3f, sizeof(f));
    dfs(0, k, 0, lim - 1);
    for (int i = 0; i < lim; ++i) write(f[i], ' ');
    return 0;
  }
  