#include <bits/stdc++.h>
const int inf = 1 << 20;
using namespace std;
int T, n, m, p, k;
int f[525000], l[525000], r[525000];
void FWT(int n) {
  for (int mid = 1; mid < (1 << n); mid <<= 1) {
    for (int j = 0; j < (1 << n); j += mid << 1) {
      for (int k = 0; k < mid; k++) {
        int tp = min(f[j + k], f[mid + j + k]), l0 = l[j + k], r0 = r[j + k],
            l1 = l[mid + j + k], r1 = r[mid + j + k];
        f[j + k] = min(tp, l1 + mid - r0);
        f[mid + j + k] = min(tp, l0 + mid - r1);
        l[j + k] = min(l0, l1 + mid), l[mid + j + k] = min(l0 + mid, l1);
        r[j + k] = max(r0, r1 + mid), r[mid + j + k] = max(r0 + mid, r1);
      }
    }
  }
}
int main() {
  scanf("%d%d", &n, &m);
  for (int S = 0; S < (1 << m); S++) l[S] = f[S] = inf, r[S] = -inf;
  for (int i = 1, w; i <= n; i++) scanf("%d", &w), l[w] = r[w] = 0;
  FWT(m);
  for (int S = 0; S < (1 << m); S++) printf("%d ", f[S]);
  puts("");
}

### Prompt

Create a solution in cpp for 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>
const int inf = 1 << 20;
using namespace std;
int T, n, m, p, k;
int f[525000], l[525000], r[525000];
void FWT(int n) {
  for (int mid = 1; mid < (1 << n); mid <<= 1) {
    for (int j = 0; j < (1 << n); j += mid << 1) {
      for (int k = 0; k < mid; k++) {
        int tp = min(f[j + k], f[mid + j + k]), l0 = l[j + k], r0 = r[j + k],
            l1 = l[mid + j + k], r1 = r[mid + j + k];
        f[j + k] = min(tp, l1 + mid - r0);
        f[mid + j + k] = min(tp, l0 + mid - r1);
        l[j + k] = min(l0, l1 + mid), l[mid + j + k] = min(l0 + mid, l1);
        r[j + k] = max(r0, r1 + mid), r[mid + j + k] = max(r0 + mid, r1);
      }
    }
  }
}
int main() {
  scanf("%d%d", &n, &m);
  for (int S = 0; S < (1 << m); S++) l[S] = f[S] = inf, r[S] = -inf;
  for (int i = 1, w; i <= n; i++) scanf("%d", &w), l[w] = r[w] = 0;
  FWT(m);
  for (int S = 0; S < (1 << m); S++) printf("%d ", f[S]);
  puts("");
}

```