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>
  inline bool upmin(T &x, T y) {
    return y <= x ? x = y, 1 : 0;
  }
  template <typename T>
  inline bool upmax(T &x, T y) {
    return x <= y ? x = y, 1 : 0;
  }
  const long double eps = 1e-9;
  const long double pi = acos(-1);
  const int oo = 1 << 30;
  const long long loo = 1ll << 60;
  const int MAXN = 1 << 20;
  const int MAXM = 4005;
  const int MAXP = 7000000;
  const int MX = 20;
  const int G = 3;
  const int mods = 998244353;
  const int Gi = (mods + 1) / G;
  const int SZ = 10;
  const int inv2 = (mods + 1) >> 1;
  const int INF = 0x3f3f3f3f;
  namespace FastIO {
  constexpr int SIZE = (1 << 21) + 1;
  int num = 0, f;
  char ibuf[SIZE], obuf[SIZE], que[65], *iS, *iT, *oS = obuf,
                                                  *oT = obuf + SIZE - 1, c;
  inline void flush() {
    fwrite(obuf, 1, oS - obuf, stdout);
    oS = obuf;
  }
  inline void putc(char c) {
    *oS++ = c;
    if (oS == oT) flush();
  }
  inline void getc(char &c) {
    for (c = (iS == iT ? (iT = ((iS = ibuf) + fread(ibuf, 1, SIZE, stdin)),
                         (iS == iT ? EOF : *iS++))
                       : *iS++);
         (c == ' ' || c == '\0' || c == '\n') && c != EOF;
         c = (iS == iT ? (iT = ((iS = ibuf) + fread(ibuf, 1, SIZE, stdin)),
                         (iS == iT ? EOF : *iS++))
                       : *iS++))
      ;
  }
  inline void reads(char *st) {
    char c;
    int n = 0;
    getc(st[++n]);
    for (c = (iS == iT ? (iT = ((iS = ibuf) + fread(ibuf, 1, SIZE, stdin)),
                         (iS == iT ? EOF : *iS++))
                       : *iS++);
         c != ' ' && c != '\0' && c != '\n' && c != EOF;
         c = (iS == iT ? (iT = ((iS = ibuf) + fread(ibuf, 1, SIZE, stdin)),
                         (iS == iT ? EOF : *iS++))
                       : *iS++))
      st[++n] = c;
    st[++n] = '\0';
  }
  template <class I>
  inline void read(I &x) {
    for (f = 1, c = (iS == iT ? (iT = ((iS = ibuf) + fread(ibuf, 1, SIZE, stdin)),
                                (iS == iT ? EOF : *iS++))
                              : *iS++);
         c < '0' || c > '9';
         c = (iS == iT ? (iT = ((iS = ibuf) + fread(ibuf, 1, SIZE, stdin)),
                         (iS == iT ? EOF : *iS++))
                       : *iS++))
      if (c == '-') f = -1;
    for (x = 0; c >= '0' && c <= '9';
         c = (iS == iT ? (iT = ((iS = ibuf) + fread(ibuf, 1, SIZE, stdin)),
                         (iS == iT ? EOF : *iS++))
                       : *iS++))
      x = (x << 3) + (x << 1) + (c & 15);
    x *= f;
  }
  template <class I>
  inline void print(I x) {
    if (x < 0) putc('-'), x = -x;
    if (!x) putc('0');
    while (x) que[++num] = x % 10 + 48, x /= 10;
    while (num) putc(que[num--]);
  }
  inline void putstr(string st) {
    for (int i = 0; i < (int)st.size(); ++i) putc(st[i]);
  }
  struct Flusher_ {
    ~Flusher_() { flush(); }
  } io_Flusher_;
  }  // namespace FastIO
  using FastIO ::getc;
  using FastIO ::print;
  using FastIO ::putc;
  using FastIO ::putstr;
  using FastIO ::read;
  using FastIO ::reads;
  queue<int> que;
  int n, K, nodenum = -1, ch[MAXN][2], a[MAXN], mn[MAXN], mx[MAXN], ans[MAXN],
            len[MAXN], Ans[MAXN];
  int Nwnode(int x) {
    int t = ++nodenum;
    len[t] = 1 << x;
    ch[t][0] = ch[t][1] = 0;
    mn[t] = INF, mx[t] = -INF, ans[t] = INF + INF;
    return nodenum;
  }
  void up(int x) {
    mn[x] = min(mn[ch[x][0]], mn[ch[x][1]] + (len[x] >> 1));
    mx[x] = max(mx[ch[x][0]], mx[ch[x][1]] + (len[x] >> 1));
    ans[x] = min(min(ans[ch[x][0]], ans[ch[x][1]]),
                 mn[ch[x][1]] - mx[ch[x][0]] + (len[x] >> 1));
  }
  void insert(int x) {
    int nw = 1;
    for (int i = K - 1; i >= 0; --i) {
      if (!ch[nw][(x >> i) & 1]) ch[nw][(x >> i) & 1] = Nwnode(i);
      nw = ch[nw][(x >> i) & 1], a[i] = nw;
    }
    mn[nw] = mx[nw] = 0;
    for (int i = 1; i < K; ++i) up(a[i]);
    up(1);
  }
  void dfs(int x, int y) {
    if (x == K) {
      Ans[y] = ans[1];
      return;
    }
    dfs(x + 1, y);
    auto flip = [&]() {
      que.push(1);
      int num = 0;
      while (!que.empty()) {
        int q = que.front();
        que.pop(), a[++num] = q;
        if (len[q] == (1 << (x + 1))) {
          swap(ch[q][0], ch[q][1]);
          continue;
        }
        if (ch[q][0]) que.push(ch[q][0]);
        if (ch[q][1]) que.push(ch[q][1]);
      }
      while (num) up(a[num--]);
    };
    flip();
    dfs(x + 1, y | (1 << x));
    flip();
  }
  signed main() {
    read(n), read(K), Nwnode(0), Nwnode(K);
    for (int i = 1, x; i <= n; ++i) read(x), insert(x);
    Ans[0] = ans[1];
    dfs(0, 0);
    for (int i = 0; i < 1 << K; ++i) print(Ans[i]), putc(' ');
    return 0;
  }
  