#include <bits/stdc++.h>
using namespace std;
const int M = 998244353;
const int maxn = 1e6 + 10;
const int inf = 1e9;
int n, k, mx[21][1 << 19], mn[21][1 << 19], ans[21][1 << 19], sz[21][1 << 19];
void Min(int &x, int y) {
  if (y < x) x = y;
}
int main() {
  scanf("%d%d", &n, &k);
  for (int i = 0; i < n; i++) {
    int x;
    scanf("%d", &x);
    sz[0][x]++;
    mx[0][x] = mn[0][x] = 0;
    ans[0][x] = inf;
  }
  for (int i = 0; i < k; i++) {
    int d = 1 << i;
    for (int j = 0; j < (1 << k); j += d * 2) {
      sz[i + 1][j] = sz[i][j] + sz[i][j + d];
      if (!sz[i + 1][j]) continue;
      if (!sz[i][j + d]) {
        for (int r = 0; r < d; r++) {
          mx[i + 1][j + r] = mx[i][j + r];
          mn[i + 1][j + r] = mn[i][j + r];
          mx[i + 1][j + r + d] = mx[i][j + r] + d;
          mn[i + 1][j + r + d] = mn[i][j + r] + d;
          ans[i + 1][j + r] = ans[i + 1][j + r + d] = ans[i][j + r];
        }
      } else if (!sz[i][j]) {
        for (int r = 0; r < d; r++) {
          mx[i + 1][j + r] = mx[i][j + r + d] + d;
          mn[i + 1][j + r] = mn[i][j + r + d] + d;
          mx[i + 1][j + r + d] = mx[i][j + r + d];
          mn[i + 1][j + r + d] = mn[i][j + r + d];
          ans[i + 1][j + r] = ans[i + 1][j + r + d] = ans[i][j + r + d];
        }
      } else {
        for (int r = 0; r < d; r++) {
          mx[i + 1][j + r] = mx[i][j + r + d] + d;
          mn[i + 1][j + r] = mn[i][j + r];
          mx[i + 1][j + r + d] = mx[i][j + r] + d;
          mn[i + 1][j + r + d] = mn[i][j + r + d];
          ans[i + 1][j + r] = ans[i + 1][j + r + d] =
              min(ans[i][j + r], ans[i][j + r + d]);
          Min(ans[i + 1][j + r], (mn[i][j + r + d] + d) - mx[i][j + r]);
          Min(ans[i + 1][j + r + d], (mn[i][j + r] + d) - mx[i][j + r + d]);
        }
      }
    }
  }
  for (int i = 0; i < (1 << k); i++) printf("%d ", ans[k][i]);
  puts("");
  return 0;
}

### Prompt

Construct a Cpp code solution to the problem outlined:

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;
const int M = 998244353;
const int maxn = 1e6 + 10;
const int inf = 1e9;
int n, k, mx[21][1 << 19], mn[21][1 << 19], ans[21][1 << 19], sz[21][1 << 19];
void Min(int &x, int y) {
  if (y < x) x = y;
}
int main() {
  scanf("%d%d", &n, &k);
  for (int i = 0; i < n; i++) {
    int x;
    scanf("%d", &x);
    sz[0][x]++;
    mx[0][x] = mn[0][x] = 0;
    ans[0][x] = inf;
  }
  for (int i = 0; i < k; i++) {
    int d = 1 << i;
    for (int j = 0; j < (1 << k); j += d * 2) {
      sz[i + 1][j] = sz[i][j] + sz[i][j + d];
      if (!sz[i + 1][j]) continue;
      if (!sz[i][j + d]) {
        for (int r = 0; r < d; r++) {
          mx[i + 1][j + r] = mx[i][j + r];
          mn[i + 1][j + r] = mn[i][j + r];
          mx[i + 1][j + r + d] = mx[i][j + r] + d;
          mn[i + 1][j + r + d] = mn[i][j + r] + d;
          ans[i + 1][j + r] = ans[i + 1][j + r + d] = ans[i][j + r];
        }
      } else if (!sz[i][j]) {
        for (int r = 0; r < d; r++) {
          mx[i + 1][j + r] = mx[i][j + r + d] + d;
          mn[i + 1][j + r] = mn[i][j + r + d] + d;
          mx[i + 1][j + r + d] = mx[i][j + r + d];
          mn[i + 1][j + r + d] = mn[i][j + r + d];
          ans[i + 1][j + r] = ans[i + 1][j + r + d] = ans[i][j + r + d];
        }
      } else {
        for (int r = 0; r < d; r++) {
          mx[i + 1][j + r] = mx[i][j + r + d] + d;
          mn[i + 1][j + r] = mn[i][j + r];
          mx[i + 1][j + r + d] = mx[i][j + r] + d;
          mn[i + 1][j + r + d] = mn[i][j + r + d];
          ans[i + 1][j + r] = ans[i + 1][j + r + d] =
              min(ans[i][j + r], ans[i][j + r + d]);
          Min(ans[i + 1][j + r], (mn[i][j + r + d] + d) - mx[i][j + r]);
          Min(ans[i + 1][j + r + d], (mn[i][j + r] + d) - mx[i][j + r + d]);
        }
      }
    }
  }
  for (int i = 0; i < (1 << k); i++) printf("%d ", ans[k][i]);
  puts("");
  return 0;
}

```