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;
  const int N = 1 << 20;
  int n, k, a[N], cnt;
  struct trie {
    int son[2], de;
  } sh[N];
  struct node {
    int l, r, ans;
  };
  vector<node> p[N];
  inline int read() {
    int f = 1, x = 0;
    char s = getchar();
    while (s < '0' || s > '9') {
      if (s == '-') f = -1;
      s = getchar();
    }
    while (s >= '0' && s <= '9') {
      x = x * 10 + s - '0';
      s = getchar();
    }
    return x * f;
  }
  void insert(int num) {
    int x = 0;
    for (int i = k - 1; i >= 0; i--) {
      int now = ((num >> i) & 1), p = sh[x].son[now];
      if (!p) p = ++cnt;
      sh[x].son[now] = p;
      sh[p].de = sh[x].de + 1;
      x = p;
    }
    p[x].push_back((node){num, num, 0x3f3f3f3f});
  }
  void upd(int x, int s) {
    int now = (int)p[s].size();
    p[x] = p[s];
    p[x].resize(now << 1);
    for (int i = now; i < (int)p[x].size(); i++) p[x][i] = p[x][i - now];
  }
  inline void pushup(int x) {
    int ls = sh[x].son[0], rs = sh[x].son[1];
    if (!ls) {
      upd(x, rs);
      return;
    }
    if (!rs) {
      upd(x, ls);
      return;
    }
    int now = (int)p[ls].size();
    p[x].resize(now << 1);
    for (int i = 0; i < now; i++) {
      p[x][i].l = p[ls][i].l;
      p[x][i].r = p[rs][i].r;
      p[x][i].ans = min(p[ls][i].ans, p[rs][i].ans);
      p[x][i].ans = min(p[x][i].ans, (p[rs][i].l ^ i) - (p[ls][i].r ^ i));
    }
    for (int i = now; i < (int)p[x].size(); i++) {
      p[x][i].l = p[rs][i - now].l;
      p[x][i].r = p[ls][i - now].r;
      p[x][i].ans = min(p[ls][i - now].ans, p[rs][i - now].ans);
      p[x][i].ans =
          min(p[x][i].ans, (p[ls][i - now].l ^ i) - (p[rs][i - now].r ^ i));
    }
  }
  void dfs(int x) {
    if (sh[x].de == k) return;
    if (sh[x].son[0]) dfs(sh[x].son[0]);
    if (sh[x].son[1]) dfs(sh[x].son[1]);
    pushup(x);
  }
  signed main() {
    n = read();
    k = read();
    for (int i = 1; i <= n; i++) insert(read());
    dfs(0);
    assert((int)p[0].size() == (1 << k));
    for (int i = 0; i < (int)p[0].size(); i++) printf("%d ", p[0][i].ans);
    printf("\n");
  }
  