#include <bits/stdc++.h>
using namespace std;
const int N = 1 << 20;
const int inf = 1e9;
int k, m, n, a[N], s[N], h[N], f[N];
int mx(int l, int r, int x) {
  if (s[r] == s[l]) return -inf;
  if (r - l == 1) return l ^ x;
  int h = l + r >> 1;
  if (x & (h - l)) {
    if (s[h] == s[l]) return mx(h, r, x);
    return mx(l, h, x);
  } else {
    if (s[h] == s[r]) return mx(l, h, x);
    return mx(h, r, x);
  }
}
int mn(int l, int r, int x) {
  if (s[r] == s[l]) return inf;
  if (r - l == 1) return l ^ x;
  int h = l + r >> 1;
  if (x & (h - l)) {
    if (s[h] == s[r]) return mn(l, h, x);
    return mn(h, r, x);
  } else {
    if (s[h] == s[l]) return mn(h, r, x);
    return mn(l, h, x);
  }
}
int main() {
  int i, j, l, x;
  scanf("%d%d", &n, &k);
  m = 1 << k;
  for (i = 1; i <= n; i++) scanf("%d", &x), a[x] = 1;
  for (i = m; i >= 0; i--) s[i] = s[i + 1] + a[i];
  for (i = 0; i < m; i++) f[i] = inf;
  for (j = k - 1; j >= 0; j--) {
    for (i = 0; i < (2 << j); i++) h[i] = inf;
    for (l = 0; l < m; l = l + (2 << j)) {
      for (i = 0; i < (1 << j); i++)
        h[i] = min(h[i],
                   mn(l + (1 << j), l + (2 << j), i) - mx(l, l + (1 << j), i));
      for (i = (1 << j); i < (2 << j); i++)
        h[i] = min(h[i],
                   mn(l, l + (1 << j), i) - mx(l + (1 << j), l + (2 << j), i));
    }
    for (i = 0; i < m; i++) f[i] = min(f[i], h[i & ((2 << j) - 1)]);
  }
  for (i = 0; i < m; i++) printf("%d ", f[i]);
  return 0;
}

### Prompt

Please create a solution in CPP to 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>
using namespace std;
const int N = 1 << 20;
const int inf = 1e9;
int k, m, n, a[N], s[N], h[N], f[N];
int mx(int l, int r, int x) {
  if (s[r] == s[l]) return -inf;
  if (r - l == 1) return l ^ x;
  int h = l + r >> 1;
  if (x & (h - l)) {
    if (s[h] == s[l]) return mx(h, r, x);
    return mx(l, h, x);
  } else {
    if (s[h] == s[r]) return mx(l, h, x);
    return mx(h, r, x);
  }
}
int mn(int l, int r, int x) {
  if (s[r] == s[l]) return inf;
  if (r - l == 1) return l ^ x;
  int h = l + r >> 1;
  if (x & (h - l)) {
    if (s[h] == s[r]) return mn(l, h, x);
    return mn(h, r, x);
  } else {
    if (s[h] == s[l]) return mn(h, r, x);
    return mn(l, h, x);
  }
}
int main() {
  int i, j, l, x;
  scanf("%d%d", &n, &k);
  m = 1 << k;
  for (i = 1; i <= n; i++) scanf("%d", &x), a[x] = 1;
  for (i = m; i >= 0; i--) s[i] = s[i + 1] + a[i];
  for (i = 0; i < m; i++) f[i] = inf;
  for (j = k - 1; j >= 0; j--) {
    for (i = 0; i < (2 << j); i++) h[i] = inf;
    for (l = 0; l < m; l = l + (2 << j)) {
      for (i = 0; i < (1 << j); i++)
        h[i] = min(h[i],
                   mn(l + (1 << j), l + (2 << j), i) - mx(l, l + (1 << j), i));
      for (i = (1 << j); i < (2 << j); i++)
        h[i] = min(h[i],
                   mn(l, l + (1 << j), i) - mx(l + (1 << j), l + (2 << j), i));
    }
    for (i = 0; i < m; i++) f[i] = min(f[i], h[i & ((2 << j) - 1)]);
  }
  for (i = 0; i < m; i++) printf("%d ", f[i]);
  return 0;
}

```