#include <bits/stdc++.h>
using namespace std;
struct node {
  int Max, Min, len, ans;
};
node merge(node l, node r) {
  node res;
  res.Max = max(l.Max, r.Max + l.len);
  res.Min = min(l.Min, r.Min + l.len);
  res.len = l.len + r.len;
  res.ans = min({l.ans, r.ans, r.Min + l.len - l.Max});
  return res;
}
vector<node> tr[((1 << 20) + 10)];
int n, k, cnt[((1 << 20) + 10)];
void build(int p, int l, int r) {
  tr[p].resize(r - l + 1);
  if (l == r) {
    tr[p][0].len = 1;
    tr[p][0].ans = 0x3f3f3f3f;
    if (cnt[l] > 0)
      tr[p][0].Max = tr[p][0].Min = 0;
    else
      tr[p][0].Max = -0x3f3f3f3f, tr[p][0].Min = 0x3f3f3f3f;
    return;
  }
  int mid = (l + r) >> 1;
  build((p << 1), l, mid);
  build((p << 1 | 1), mid + 1, r);
  for (int i = (0); i <= (mid - l); i++) {
    tr[p][i] = merge(tr[(p << 1)][i], tr[(p << 1 | 1)][i]);
    tr[p][i + (mid - l + 1)] = merge(tr[(p << 1 | 1)][i], tr[(p << 1)][i]);
  }
}
int main() {
  scanf("%d%d", &n, &k);
  for (int i = (1); i <= (n); i++) {
    int x;
    scanf("%d", &x);
    cnt[x]++;
  }
  build(1, 0, (1 << k) - 1);
  for (int i = (0); i <= ((1 << k) - 1); i++) {
    printf("%d ", tr[1][i].ans);
  }
  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;
struct node {
  int Max, Min, len, ans;
};
node merge(node l, node r) {
  node res;
  res.Max = max(l.Max, r.Max + l.len);
  res.Min = min(l.Min, r.Min + l.len);
  res.len = l.len + r.len;
  res.ans = min({l.ans, r.ans, r.Min + l.len - l.Max});
  return res;
}
vector<node> tr[((1 << 20) + 10)];
int n, k, cnt[((1 << 20) + 10)];
void build(int p, int l, int r) {
  tr[p].resize(r - l + 1);
  if (l == r) {
    tr[p][0].len = 1;
    tr[p][0].ans = 0x3f3f3f3f;
    if (cnt[l] > 0)
      tr[p][0].Max = tr[p][0].Min = 0;
    else
      tr[p][0].Max = -0x3f3f3f3f, tr[p][0].Min = 0x3f3f3f3f;
    return;
  }
  int mid = (l + r) >> 1;
  build((p << 1), l, mid);
  build((p << 1 | 1), mid + 1, r);
  for (int i = (0); i <= (mid - l); i++) {
    tr[p][i] = merge(tr[(p << 1)][i], tr[(p << 1 | 1)][i]);
    tr[p][i + (mid - l + 1)] = merge(tr[(p << 1 | 1)][i], tr[(p << 1)][i]);
  }
}
int main() {
  scanf("%d%d", &n, &k);
  for (int i = (1); i <= (n); i++) {
    int x;
    scanf("%d", &x);
    cnt[x]++;
  }
  build(1, 0, (1 << k) - 1);
  for (int i = (0); i <= ((1 << k) - 1); i++) {
    printf("%d ", tr[1][i].ans);
  }
  return 0;
}

```