#include <bits/stdc++.h>
const int N = 20;
using namespace std;
int f[N][1 << N], X[N][1 << N], I[N][1 << N], n, k, a[1 << N];
int main() {
  ios::sync_with_stdio(false);
  cin >> n >> k;
  memset(X, -2, sizeof(X));
  memset(f, 1, sizeof(f));
  memset(I, 1, sizeof(I));
  for (int i = 1; i <= n; i++) cin >> a[i], X[0][a[i]] = I[0][a[i]] = 0;
  for (int i = 1; i <= k; i++)
    for (int j = 0, p = 1 << i - 1; j < 1 << k; j++)
      if (j >> (i - 1) & 1) {
        int x = j, y = j ^ p;
        f[i][x] = f[i][y] = min(f[i - 1][x], f[i - 1][y]);
        f[i][x] = min(f[i][x], I[i - 1][y] - X[i - 1][x] + p);
        f[i][y] = min(f[i][y], I[i - 1][x] - X[i - 1][y] + p);
        X[i][x] = max(X[i - 1][x], X[i - 1][y] + p);
        I[i][x] = min(I[i - 1][x], I[i - 1][y] + p);
        X[i][y] = max(X[i - 1][y], X[i - 1][x] + p);
        I[i][y] = min(I[i - 1][y], I[i - 1][x] + p);
      }
  for (int i = 0; i < 1 << k; i++) cout << f[k][i] << ' ';
}

### Prompt

Develop a solution in cpp 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>
const int N = 20;
using namespace std;
int f[N][1 << N], X[N][1 << N], I[N][1 << N], n, k, a[1 << N];
int main() {
  ios::sync_with_stdio(false);
  cin >> n >> k;
  memset(X, -2, sizeof(X));
  memset(f, 1, sizeof(f));
  memset(I, 1, sizeof(I));
  for (int i = 1; i <= n; i++) cin >> a[i], X[0][a[i]] = I[0][a[i]] = 0;
  for (int i = 1; i <= k; i++)
    for (int j = 0, p = 1 << i - 1; j < 1 << k; j++)
      if (j >> (i - 1) & 1) {
        int x = j, y = j ^ p;
        f[i][x] = f[i][y] = min(f[i - 1][x], f[i - 1][y]);
        f[i][x] = min(f[i][x], I[i - 1][y] - X[i - 1][x] + p);
        f[i][y] = min(f[i][y], I[i - 1][x] - X[i - 1][y] + p);
        X[i][x] = max(X[i - 1][x], X[i - 1][y] + p);
        I[i][x] = min(I[i - 1][x], I[i - 1][y] + p);
        X[i][y] = max(X[i - 1][y], X[i - 1][x] + p);
        I[i][y] = min(I[i - 1][y], I[i - 1][x] + p);
      }
  for (int i = 0; i < 1 << k; i++) cout << f[k][i] << ' ';
}

```