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. 
#include <bits/stdc++.h>
template <typename _Tp>
void read(_Tp &x) {
  char ch(getchar());
  bool f(false);
  while (!isdigit(ch)) f |= ch == 45, ch = getchar();
  x = ch & 15, ch = getchar();
  while (isdigit(ch)) x = x * 10 + (ch & 15), ch = getchar();
  if (f) x = -x;
}
template <typename _Tp, typename... Args>
void read(_Tp &t, Args &...args) {
  read(t);
  read(args...);
}
template <typename _Tp>
inline _Tp abs(const _Tp &a) {
  return a >= 0 ? a : -a;
}
const int N = (1 << 19) + 5;
int ch[N << 1][2], nc = 1;
std::vector<int> a[N << 1];
void ins(int x) {
  int u = 1;
  for (int i = 19; i >= 0; --i) {
    int tmp = x >> i & 1;
    if (!ch[u][tmp]) ch[u][tmp] = ++nc;
    u = ch[u][tmp];
  }
  a[u].push_back(x);
}
std::vector<std::pair<int, int> > v[20];
void dfs(int x, int dep) {
  if (!x) return;
  dfs(ch[x][0], dep - 1), dfs(ch[x][1], dep - 1);
  for (auto A : a[ch[x][0]])
    for (auto B : a[ch[x][1]]) v[dep].push_back({A, B});
  if (((int)a[ch[x][0]].size()) < 2)
    for (auto it : a[ch[x][0]]) a[x].push_back(it);
  if (((int)a[ch[x][1]].size()) < 2)
    for (auto it : a[ch[x][1]]) a[x].push_back(it);
}
int ans[N], res[N];
int main() {
  int n, k;
  read(n, k);
  for (int i = 1, x; i <= n; ++i) read(x), ins(x);
  dfs(1, 19);
  memset(res, 63, sizeof(res));
  for (int i = 0; i < 20; ++i)
    if (!v[i].empty()) {
      memset(ans, 63, sizeof(ans));
      for (int x = 0; x < 1 << (i + 1); ++x)
        for (auto [a, b] : v[i])
          ans[x] = std::min(ans[x], abs((a ^ x) - (b ^ x)));
      int M = (1 << (i + 1)) - 1;
      for (int i = 0; i < 1 << k; ++i) res[i] = std::min(res[i], ans[i & M]);
    }
  for (int i = 0; i < 1 << k; ++i)
    printf("%d%c", res[i], " \n"[i + 1 == (1 << k)]);
  return 0;
}
