#include <bits/stdc++.h>
using namespace std;
const int MX = (1 << 20);
int n, m, num, i, j, k, cur, a[MX], b[MX * 21][2], ans[21][MX];
int lfs(int l, int i, int msk) {
  int cur = ((msk >> l) & 1);
  if (b[i][cur]) return lfs(l - 1, b[i][cur], msk);
  if (b[i][cur ^ 1]) return lfs(l - 1, b[i][cur ^ 1], msk) + (1 << l);
  return 0;
}
int rfs(int l, int i, int msk) {
  int cur = ((msk >> l) & 1);
  if (b[i][cur ^ 1]) return rfs(l - 1, b[i][cur ^ 1], msk) + (1 << l);
  if (b[i][cur]) return rfs(l - 1, b[i][cur], msk);
  return 0;
}
void dfs(int l, int i) {
  if (b[i][0] && b[i][1]) {
    int msk = 0, cur;
    for (; msk < (1 << l); msk++) {
      int ll = lfs(l - 1, b[i][1], msk);
      int rr = rfs(l - 1, b[i][0], msk);
      ans[l][msk] = min(ans[l][msk], (1 << l) + ll - rr);
    }
    for (; msk < 2 * (1 << l); msk++) {
      int ll = lfs(l - 1, b[i][0], msk);
      int rr = rfs(l - 1, b[i][1], msk);
      ans[l][msk] = min(ans[l][msk], (1 << l) + ll - rr);
    }
  }
  if (b[i][0]) dfs(l - 1, b[i][0]);
  if (b[i][1]) dfs(l - 1, b[i][1]);
}
int main() {
  scanf("%d%d", &n, &m);
  for (num = i = 0; i < n; i++) {
    scanf("%d", &a[i]);
    for (j = k = 0; j < m; j++) {
      cur = ((a[i] >> (m - 1 - j)) & 1);
      if (b[k][cur] == 0) b[k][cur] = ++num;
      k = b[k][cur];
    }
  }
  for (j = 0; j < m; j++)
    for (i = 0; i < (1 << (j + 1)); i++) ans[j][i] = (1 << m);
  dfs(m - 1, 0);
  for (j = 0; j + 1 < m; j++)
    for (i = 0; i < (1 << (j + 1)); i++)
      for (k = i; k < (1 << m); k += (1 << (j + 1)))
        ans[m - 1][k] = min(ans[m - 1][k], ans[j][i]);
  for (i = 0; i < (1 << m); i++) printf("%d ", ans[m - 1][i]);
  return 0;
}

### 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;
const int MX = (1 << 20);
int n, m, num, i, j, k, cur, a[MX], b[MX * 21][2], ans[21][MX];
int lfs(int l, int i, int msk) {
  int cur = ((msk >> l) & 1);
  if (b[i][cur]) return lfs(l - 1, b[i][cur], msk);
  if (b[i][cur ^ 1]) return lfs(l - 1, b[i][cur ^ 1], msk) + (1 << l);
  return 0;
}
int rfs(int l, int i, int msk) {
  int cur = ((msk >> l) & 1);
  if (b[i][cur ^ 1]) return rfs(l - 1, b[i][cur ^ 1], msk) + (1 << l);
  if (b[i][cur]) return rfs(l - 1, b[i][cur], msk);
  return 0;
}
void dfs(int l, int i) {
  if (b[i][0] && b[i][1]) {
    int msk = 0, cur;
    for (; msk < (1 << l); msk++) {
      int ll = lfs(l - 1, b[i][1], msk);
      int rr = rfs(l - 1, b[i][0], msk);
      ans[l][msk] = min(ans[l][msk], (1 << l) + ll - rr);
    }
    for (; msk < 2 * (1 << l); msk++) {
      int ll = lfs(l - 1, b[i][0], msk);
      int rr = rfs(l - 1, b[i][1], msk);
      ans[l][msk] = min(ans[l][msk], (1 << l) + ll - rr);
    }
  }
  if (b[i][0]) dfs(l - 1, b[i][0]);
  if (b[i][1]) dfs(l - 1, b[i][1]);
}
int main() {
  scanf("%d%d", &n, &m);
  for (num = i = 0; i < n; i++) {
    scanf("%d", &a[i]);
    for (j = k = 0; j < m; j++) {
      cur = ((a[i] >> (m - 1 - j)) & 1);
      if (b[k][cur] == 0) b[k][cur] = ++num;
      k = b[k][cur];
    }
  }
  for (j = 0; j < m; j++)
    for (i = 0; i < (1 << (j + 1)); i++) ans[j][i] = (1 << m);
  dfs(m - 1, 0);
  for (j = 0; j + 1 < m; j++)
    for (i = 0; i < (1 << (j + 1)); i++)
      for (k = i; k < (1 << m); k += (1 << (j + 1)))
        ans[m - 1][k] = min(ans[m - 1][k], ans[j][i]);
  for (i = 0; i < (1 << m); i++) printf("%d ", ans[m - 1][i]);
  return 0;
}

```