#include <bits/stdc++.h>
using namespace std;
const int N = 530005;
int n, k, i, t, m, j, len, a[N], ch[N * 20][2], p, tot, mn[N], mx[N], ans[N];
int as[20][N];
int Findmn(int i, int d, int x) {
  if (d == -1) return 0;
  if (ch[i][x >> d & 1])
    return Findmn(ch[i][x >> d & 1], d - 1, x);
  else
    return Findmn(ch[i][(x >> d & 1) ^ 1], d - 1, x) + (1 << d);
}
int Findmx(int i, int d, int x) {
  if (d == -1) return 0;
  if (ch[i][(x >> d & 1) ^ 1])
    return Findmx(ch[i][(x >> d & 1) ^ 1], d - 1, x) + (1 << d);
  else
    return Findmx(ch[i][x >> d & 1], d - 1, x);
}
void dfs(int i, int d) {
  if (!i || d == -1) return;
  if (ch[i][0] && ch[i][1]) {
    for (j = 0; j < (1 << d); ++j) {
      as[d + 1][j] = min(as[d + 1][j], (1 << d) - Findmx(ch[i][0], d - 1, j) +
                                           Findmn(ch[i][1], d - 1, j));
      as[d + 1][j + (1 << d)] =
          min(as[d + 1][j + (1 << d)], (1 << d) - Findmx(ch[i][1], d - 1, j) +
                                           Findmn(ch[i][0], d - 1, j));
    }
  }
  dfs(ch[i][0], d - 1);
  dfs(ch[i][1], d - 1);
}
int main() {
  scanf("%d %d", &n, &k);
  tot = 1;
  for (i = 1; i <= n; ++i) {
    scanf("%d", &a[i]);
    p = 1;
    for (j = k - 1; j >= 0; --j) {
      if (!ch[p][a[i] >> j & 1]) ch[p][a[i] >> j & 1] = ++tot;
      p = ch[p][a[i] >> j & 1];
    }
  }
  for (i = 0; i <= k; ++i)
    for (j = 0; j < (1 << i); ++j) as[i][j] = 1 << 30;
  dfs(1, k - 1);
  for (i = 0; i < (1 << k); ++i) ans[i] = 1 << 30;
  for (i = 0; i <= k; ++i)
    for (j = 0; j < (1 << k); ++j) ans[j] = min(ans[j], as[i][j % (1 << i)]);
  for (i = 0; i < (1 << k); ++i) printf("%d ", ans[i]);
}

### 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 N = 530005;
int n, k, i, t, m, j, len, a[N], ch[N * 20][2], p, tot, mn[N], mx[N], ans[N];
int as[20][N];
int Findmn(int i, int d, int x) {
  if (d == -1) return 0;
  if (ch[i][x >> d & 1])
    return Findmn(ch[i][x >> d & 1], d - 1, x);
  else
    return Findmn(ch[i][(x >> d & 1) ^ 1], d - 1, x) + (1 << d);
}
int Findmx(int i, int d, int x) {
  if (d == -1) return 0;
  if (ch[i][(x >> d & 1) ^ 1])
    return Findmx(ch[i][(x >> d & 1) ^ 1], d - 1, x) + (1 << d);
  else
    return Findmx(ch[i][x >> d & 1], d - 1, x);
}
void dfs(int i, int d) {
  if (!i || d == -1) return;
  if (ch[i][0] && ch[i][1]) {
    for (j = 0; j < (1 << d); ++j) {
      as[d + 1][j] = min(as[d + 1][j], (1 << d) - Findmx(ch[i][0], d - 1, j) +
                                           Findmn(ch[i][1], d - 1, j));
      as[d + 1][j + (1 << d)] =
          min(as[d + 1][j + (1 << d)], (1 << d) - Findmx(ch[i][1], d - 1, j) +
                                           Findmn(ch[i][0], d - 1, j));
    }
  }
  dfs(ch[i][0], d - 1);
  dfs(ch[i][1], d - 1);
}
int main() {
  scanf("%d %d", &n, &k);
  tot = 1;
  for (i = 1; i <= n; ++i) {
    scanf("%d", &a[i]);
    p = 1;
    for (j = k - 1; j >= 0; --j) {
      if (!ch[p][a[i] >> j & 1]) ch[p][a[i] >> j & 1] = ++tot;
      p = ch[p][a[i] >> j & 1];
    }
  }
  for (i = 0; i <= k; ++i)
    for (j = 0; j < (1 << i); ++j) as[i][j] = 1 << 30;
  dfs(1, k - 1);
  for (i = 0; i < (1 << k); ++i) ans[i] = 1 << 30;
  for (i = 0; i <= k; ++i)
    for (j = 0; j < (1 << k); ++j) ans[j] = min(ans[j], as[i][j % (1 << i)]);
  for (i = 0; i < (1 << k); ++i) printf("%d ", ans[i]);
}

```