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>
  bool chkmax(T &x, T y) {
    return x < y ? x = y, true : false;
  }
  template <typename T>
  bool chkmin(T &x, T y) {
    return x > y ? x = y, true : false;
  }
  int readint() {
    int x = 0, f = 1;
    char ch = getchar();
    while (ch < '0' || ch > '9') {
      if (ch == '-') f = -1;
      ch = getchar();
    }
    while (ch >= '0' && ch <= '9') {
      x = x * 10 + ch - '0';
      ch = getchar();
    }
    return x * f;
  }
  int n, k, ncnt;
  int ch[10000005][2], mina[10000005], maxa[10000005], res[10000005],
      ans[1100005];
  vector<int> vec[25];
  void insert(int x) {
    int p = 0;
    for (int i = k - 1; i >= 0; i--) {
      int t = (x >> i) & 1;
      if (!ch[p][t]) {
        ch[p][t] = ++ncnt;
        if (i) vec[i - 1].push_back(ncnt);
      }
      p = ch[p][t];
    }
  }
  void update(int u, int dep, int lim) {
    if (dep == lim) {
      mina[u] ^= 1 << lim, maxa[u] ^= 1 << lim;
      return;
    }
    res[u] = 1 << 30;
    if (ch[u][0]) update(ch[u][0], dep - 1, lim), chkmin(res[u], res[ch[u][0]]);
    if (ch[u][1]) update(ch[u][1], dep - 1, lim), chkmin(res[u], res[ch[u][1]]);
    if (ch[u][0] && ch[u][1]) chkmin(res[u], mina[ch[u][1]] - maxa[ch[u][0]]);
    mina[u] = 1 << 30, maxa[u] = 0;
    if (ch[u][0])
      chkmin(mina[u], mina[ch[u][0]]), chkmax(maxa[u], maxa[ch[u][0]]);
    if (ch[u][1])
      chkmin(mina[u], mina[ch[u][1]]), chkmax(maxa[u], maxa[ch[u][1]]);
  }
  void dfs(int u, int now, int dep) {
    if (dep < 0) {
      mina[u] = maxa[u] = now, res[u] = 1 << 30;
      return;
    }
    res[u] = 1 << 30;
    if (ch[u][0]) dfs(ch[u][0], now, dep - 1), chkmin(res[u], res[ch[u][0]]);
    if (ch[u][1])
      dfs(ch[u][1], now ^ (1 << dep), dep - 1), chkmin(res[u], res[ch[u][1]]);
    if (ch[u][0] && ch[u][1]) chkmin(res[u], mina[ch[u][1]] - maxa[ch[u][0]]);
    mina[u] = 1 << 30, maxa[u] = 0;
    if (ch[u][0])
      chkmin(mina[u], mina[ch[u][0]]), chkmax(maxa[u], maxa[ch[u][0]]);
    if (ch[u][1])
      chkmin(mina[u], mina[ch[u][1]]), chkmax(maxa[u], maxa[ch[u][1]]);
  }
  void work(int dep, int now) {
    if (dep == k) {
      ans[now] = res[0];
      return;
    }
    work(dep + 1, now);
    now ^= 1 << dep;
    for (auto r : vec[dep]) swap(ch[r][0], ch[r][1]);
    update(0, k, dep);
    work(dep + 1, now);
    for (auto r : vec[dep]) swap(ch[r][0], ch[r][1]);
  }
  int main() {
    n = readint();
    k = readint();
    vec[k - 1].push_back(0);
    for (int i = 1; i <= n; i++) insert(readint());
    dfs(0, 0, k - 1);
    work(0, 0);
    for (int i = 0; i < (1 << k); i++) printf("%d ", ans[i]);
    printf("\n");
    return 0;
  }
  