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>
  void sc(int &x) { scanf("%d", &x); }
  void sc(int &x, int &y) { scanf("%d%d", &x, &y); }
  void sc(int &x, int &y, int &z) { scanf("%d%d%d", &x, &y, &z); }
  void sc(long long &x) { scanf("%lld", &x); }
  void sc(long long &x, long long &y) { scanf("%lld%lld", &x, &y); }
  void sc(long long &x, long long &y, long long &z) {
    scanf("%lld%lld%lld", &x, &y, &z);
  }
  void sc(char *x) { scanf("%s", x); }
  void sc(char *x, char *y) { scanf("%s%s", x, y); }
  void sc(char *x, char *y, char *z) { scanf("%s%s%s", x, y, z); }
  void out(int x) { printf("%d\n", x); }
  void out(long long x) { printf("%lld\n", x); }
  void out(int x, int y) { printf("%d %d\n", x, y); }
  void out(long long x, long long y) { printf("%lld %lld\n", x, y); }
  void out(int x, int y, int z) { printf("%d %d %d\n", x, y, z); }
  void out(long long x, long long y, long long z) {
    printf("%lld %lld %lld\n", x, y, z);
  }
  void ast(long long x, long long l, long long r) { assert(x >= l && x <= r); }
  using namespace std;
  const int N = 1e6 + 5, mod = 1e9 + 7;
  int n, k, rt = 1, tot = 1, t[N * 10][2], mn[N * 10], mx[N * 10];
  vector<int> vc[25];
  int ans[25];
  void ins(int x) {
    int now = rt;
    for (int j = k - 1; j >= 0; j--) {
      int p = x >> j & 1;
      if (p) x ^= (1 << j);
      if (!t[now][p]) {
        t[now][p] = ++tot;
        mn[tot] = mx[tot] = x;
      }
      now = t[now][p];
      mn[now] = min(mn[now], x);
      mx[now] = max(mx[now], x);
    }
  }
  void dfs(int u, int bit) {
    if (!u) return;
    vc[bit].push_back(u);
    if (t[u][0] && t[u][1])
      ans[bit] = min(ans[bit], mn[t[u][1]] + (1 << bit) - mx[t[u][0]]);
    if (bit) {
      dfs(t[u][0], bit - 1);
      dfs(t[u][1], bit - 1);
    }
  }
  int answer[N], rev[N];
  void update(int u, int bit) {
    if (!u) return;
    if (t[u][0])
      mn[u] = mn[t[u][0]];
    else
      mn[u] = (1 << bit) + mn[t[u][1]];
    if (t[u][1])
      mx[u] = (1 << bit) + mx[t[u][1]];
    else
      mx[u] = mx[t[u][0]];
  }
  int code[N];
  void GrayCode(int k) {
    code[0] = 0;
    code[1] = 1;
    int len = 2;
    for (int j = 2; j <= k; j++) {
      for (int l = len - 1, r = len; l >= 0; l--, r++)
        code[r] = code[l] | (1 << (j - 1));
      len *= 2;
    }
  }
  void Rev(int bit) {
    ans[bit] = 998244353;
    for (int u : vc[bit]) {
      swap(t[u][0], t[u][1]);
      if (t[u][0] && t[u][1])
        ans[bit] = min(ans[bit], mn[t[u][1]] + (1 << bit) - mx[t[u][0]]);
    }
    for (int i = bit + 1; i < k; i++) {
      ans[i] = 998244353;
      for (int u : vc[i]) {
        update(t[u][0], i - 1);
        update(t[u][1], i - 1);
        if (t[u][0] && t[u][1])
          ans[i] = min(ans[i], mn[t[u][1]] + (1 << i) - mx[t[u][0]]);
      }
    }
  }
  void sol(int cas) {
    sc(n, k);
    for (int i = 1; i <= n; i++) {
      int x;
      sc(x);
      ins(x);
    }
    for (int i = 0; i <= k - 1; i++) ans[i] = 998244353;
    dfs(rt, k - 1);
    GrayCode(k);
    int up = 1 << k;
    for (int i = 1; i < up; i++)
      rev[i] = (rev[i >> 1] >> 1) | ((i & 1) << (k - 1));
    for (int i = 0; i < up; i++) {
      int res = 998244353;
      for (int j = 0; j <= k - 1; j++) res = min(res, ans[j]);
      answer[rev[code[i]]] = res;
      if (i + 1 < up) {
        int x = rev[code[i]] ^ rev[code[i + 1]];
        for (int j = 0; j < k; j++)
          if (x >> j & 1) Rev(j);
      }
    }
    for (int i = 0; i <= up - 1; i++)
      printf(i == up - 1 ? "%d\n" : "%d ", answer[i]);
  }
  int main() {
    srand(time(0));
    int t = 1, cas = 0;
    while (t--) {
      sol(++cas);
    }
  }
  