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 = 2000005;
  int n, k;
  int a[N];
  int ch[N][2], siz[N], tot = 1;
  int num[21];
  int ans[N], nowans[N], now[N * 10];
  int dongtot = 0;
  bool Fa = false;
  void Ins(int x) {
    for (int i = 20; i >= 1; i--) {
      num[i] = (x & 1);
      x >>= 1;
    }
    int now = 1;
    siz[now]++;
    for (int i = 1; i <= 20; i++) {
      if (!ch[now][num[i]]) ch[now][num[i]] = (++tot);
      now = ch[now][num[i]];
      siz[now]++;
    }
    if (siz[now] >= 2) {
      Fa = true;
    }
  }
  inline int Min(int A, int B) { return (A > B) ? B : A; }
  int Qmin(int posi, int x, int qian) {
    if (!qian) return 0;
    int topos = ((x & qian) ? 1 : 0);
    if (ch[posi][topos])
      return Qmin(ch[posi][topos], x, qian >> 1);
    else
      return (qian) + Qmin(ch[posi][!topos], x, qian >> 1);
  }
  int Qmax(int posi, int x, int qian) {
    if (!qian) return 0;
    int topos = ((x & qian) ? 1 : 0);
    if (ch[posi][!topos])
      return (qian) + Qmax(ch[posi][!topos], x, qian >> 1);
    else
      return Qmax(ch[posi][topos], x, qian >> 1);
  }
  void Solve(int posi, int qian) {
    if (siz[posi] < 2) {
      for (int i = 0; i < qian; i++) nowans[i] = 2147483647;
      return;
    }
    int hahaha = (qian >> 1);
    int L = dongtot + 1;
    dongtot += qian;
    for (int i = 0; i < qian; i++) now[L + i] = 2147483647;
    Solve(ch[posi][0], hahaha);
    for (int i = 0; i < (hahaha); i++) {
      now[L + i] = Min(now[L + i], nowans[i]);
      now[L + i + (hahaha)] = Min(now[L + i + (hahaha)], nowans[i]);
    }
    Solve(ch[posi][1], hahaha);
    for (int i = 0; i < (hahaha); i++) {
      now[L + i] = Min(now[L + i], nowans[i]);
      now[L + i + (hahaha)] = Min(now[L + i + (hahaha)], nowans[i]);
    }
    if (!ch[posi][0] || !ch[posi][1]) {
      for (int i = 0; i < qian; i++) {
        nowans[i] = now[L + i];
        ans[i] = Min(ans[i], nowans[i]);
      }
      return;
    }
    for (int i = 0; i < (hahaha); i++) {
      int Md1 = Qmin(ch[posi][1], i, hahaha >> 1);
      int Md2 = Qmax(ch[posi][0], i, hahaha >> 1);
      now[L + i] = Min(now[L + i], Md1 - Md2 + (hahaha));
    }
    for (int i = (hahaha); i < qian; i++) {
      int Md1 = Qmin(ch[posi][0], i ^ (hahaha), hahaha >> 1);
      int Md2 = Qmax(ch[posi][1], i ^ (hahaha), hahaha >> 1);
      now[L + i] = Min(now[L + i], Md1 - Md2 + (hahaha));
    }
    for (int i = 0; i < qian; i++) {
      nowans[i] = now[L + i];
      ans[i] = Min(ans[i], nowans[i]);
    }
  }
  int main() {
    scanf("%d%d", &n, &k);
    for (int i = 1; i <= n; i++) {
      scanf("%d", &a[i]);
      Ins(a[i]);
    }
    if (Fa) {
      for (int i = 0; i < (1 << k); i++) printf("0 ");
      return 0;
    }
    for (int i = 0; i < (1 << k); i++) ans[i] = 2147483647;
    Solve(1, (1 << 20));
    for (int i = 0; i < (1 << k); i++) printf("%d ", ans[i]);
  }
  