#include <bits/stdc++.h>
using namespace std;
int main() {
  long long n, k;
  cin >> n >> k;
  vector<long long> act(1 << k, 0);
  for (long long i = 0; i < n; ++i) {
    long long t;
    cin >> t;
    act[t] = 1;
  }
  function<vector<vector<long long>>(long long, long long, long long)> solve =
      [&](long long l, long long r, long long ind) {
        if (l + 1 == r)
          return vector<vector<long long>>{{LLONG_MAX / 2 * (!act[l])},
                                           {-LLONG_MAX / 2 * (!act[l])},
                                           {LLONG_MAX / 2}};
        else {
          long long mid = (l + r) / 2;
          long long le = r - l;
          vector<vector<long long>> ret = {
              vector<long long>(le, LLONG_MAX / 2),
              vector<long long>(le, -LLONG_MAX / 2),
              vector<long long>(le, LLONG_MAX / 2)};
          auto q1 = solve(l, mid, ind - 1), q2 = solve(mid, r, ind - 1);
          for (long long i = 0; i < le / 2; ++i) {
            ret[0][i] = min(q1[0][i], q2[0][i] + (1 << ind));
            ret[0][i + le / 2] = min(q2[0][i], q1[0][i] + (1 << ind));
            ret[1][i + le / 2] = max(q1[1][i] + (1 << ind), q2[1][i]);
            ret[1][i] = max(q2[1][i] + (1 << ind), q1[1][i]);
            ret[2][i] = min(q1[2][i], q2[2][i]);
            ret[2][i + le / 2] = min(q1[2][i], q2[2][i]);
          }
          for (long long i = 0; i < le / 2; ++i)
            ret[2][i] = min(ret[2][i], q2[0][i] + (1 << ind) - q1[1][i]);
          for (long long i = 0; i < le / 2; ++i)
            ret[2][i + le / 2] =
                min(ret[2][i + le / 2], q1[0][i] + (1 << ind) - q2[1][i]);
          return ret;
        }
      };
  vector<long long> ans = solve(0, 1 << k, k - 1)[2];
  for (long long i = 0; i < 1 << k; ++i) cout << ans[i] << " ";
  cout << endl;
}

### Prompt

Please provide a Cpp coded solution to the problem described below:

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;
int main() {
  long long n, k;
  cin >> n >> k;
  vector<long long> act(1 << k, 0);
  for (long long i = 0; i < n; ++i) {
    long long t;
    cin >> t;
    act[t] = 1;
  }
  function<vector<vector<long long>>(long long, long long, long long)> solve =
      [&](long long l, long long r, long long ind) {
        if (l + 1 == r)
          return vector<vector<long long>>{{LLONG_MAX / 2 * (!act[l])},
                                           {-LLONG_MAX / 2 * (!act[l])},
                                           {LLONG_MAX / 2}};
        else {
          long long mid = (l + r) / 2;
          long long le = r - l;
          vector<vector<long long>> ret = {
              vector<long long>(le, LLONG_MAX / 2),
              vector<long long>(le, -LLONG_MAX / 2),
              vector<long long>(le, LLONG_MAX / 2)};
          auto q1 = solve(l, mid, ind - 1), q2 = solve(mid, r, ind - 1);
          for (long long i = 0; i < le / 2; ++i) {
            ret[0][i] = min(q1[0][i], q2[0][i] + (1 << ind));
            ret[0][i + le / 2] = min(q2[0][i], q1[0][i] + (1 << ind));
            ret[1][i + le / 2] = max(q1[1][i] + (1 << ind), q2[1][i]);
            ret[1][i] = max(q2[1][i] + (1 << ind), q1[1][i]);
            ret[2][i] = min(q1[2][i], q2[2][i]);
            ret[2][i + le / 2] = min(q1[2][i], q2[2][i]);
          }
          for (long long i = 0; i < le / 2; ++i)
            ret[2][i] = min(ret[2][i], q2[0][i] + (1 << ind) - q1[1][i]);
          for (long long i = 0; i < le / 2; ++i)
            ret[2][i + le / 2] =
                min(ret[2][i + le / 2], q1[0][i] + (1 << ind) - q2[1][i]);
          return ret;
        }
      };
  vector<long long> ans = solve(0, 1 << k, k - 1)[2];
  for (long long i = 0; i < 1 << k; ++i) cout << ans[i] << " ";
  cout << endl;
}

```