#include <bits/stdc++.h>
using namespace std;
using ll = long long;
template <typename T>
bool chmin(T& x, T y) {
  return y < x ? (x = y, true) : false;
}
template <typename T>
bool chmax(T& x, T y) {
  return x < y ? (x = y, true) : false;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int n, k;
  cin >> n >> k;
  const int inf = 1 << (k + 1);
  vector<int> ans(1 << k, inf), mx(1 << k, -inf), mn(1 << k, +inf);
  for (int i = 0; i < n; ++i) {
    int a;
    cin >> a;
    mx[a] = mn[a] = 0;
  }
  for (int l = 0; l < k; ++l) {
    for (int mask = 0; mask < (1 << k); ++mask) {
      if (mask >> l & 1) continue;
      int nmask = mask | 1 << l;
      int cur = min(ans[mask], ans[nmask]);
      chmin(ans[mask], min(cur, (1 << l) + mn[nmask] - mx[mask]));
      chmin(ans[nmask], min(cur, (1 << l) + mn[mask] - mx[nmask]));
      int mmx[2] = {mx[mask], mx[nmask]};
      int mmn[2] = {mn[mask], mn[nmask]};
      for (int t : {0, 1}) {
        chmin(mn[mask | t << l], mmn[t ^ 1] + (1 << l));
        chmax(mx[mask | t << l], mmx[t ^ 1] + (1 << l));
      }
    }
  }
  for (int mask = 0; mask < (1 << k); ++mask) {
    cout << ans[mask] << ' ';
  }
  cout << '\n';
  exit(0);
}

### Prompt

Your challenge is to write a Cpp solution to the following problem:

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. 

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
using ll = long long;
template <typename T>
bool chmin(T& x, T y) {
  return y < x ? (x = y, true) : false;
}
template <typename T>
bool chmax(T& x, T y) {
  return x < y ? (x = y, true) : false;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int n, k;
  cin >> n >> k;
  const int inf = 1 << (k + 1);
  vector<int> ans(1 << k, inf), mx(1 << k, -inf), mn(1 << k, +inf);
  for (int i = 0; i < n; ++i) {
    int a;
    cin >> a;
    mx[a] = mn[a] = 0;
  }
  for (int l = 0; l < k; ++l) {
    for (int mask = 0; mask < (1 << k); ++mask) {
      if (mask >> l & 1) continue;
      int nmask = mask | 1 << l;
      int cur = min(ans[mask], ans[nmask]);
      chmin(ans[mask], min(cur, (1 << l) + mn[nmask] - mx[mask]));
      chmin(ans[nmask], min(cur, (1 << l) + mn[mask] - mx[nmask]));
      int mmx[2] = {mx[mask], mx[nmask]};
      int mmn[2] = {mn[mask], mn[nmask]};
      for (int t : {0, 1}) {
        chmin(mn[mask | t << l], mmn[t ^ 1] + (1 << l));
        chmax(mx[mask | t << l], mmx[t ^ 1] + (1 << l));
      }
    }
  }
  for (int mask = 0; mask < (1 << k); ++mask) {
    cout << ans[mask] << ' ';
  }
  cout << '\n';
  exit(0);
}

```