#include <bits/stdc++.h>
using namespace std;
const long long maxn = 1 << 19;
long long n, k;
long long ans[maxn], mx[maxn], mn[maxn];
bool occ[maxn];
signed main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cin >> n >> k;
  for (long long i = 0; i < (long long)(1 << k); ++i)
    mx[i] = -1e9, mn[i] = 1e9, ans[i] = 2e9;
  for (long long i = 0; i < (long long)(n); ++i) {
    long long u;
    cin >> u;
    occ[u] = 1, mx[u] = u, mn[u] = u;
  }
  for (long long i = 0; i < k; ++i) {
    for (long long j = 0; j < (1 << k); j += (1 << i + 1)) {
      for (long long t = 0; t < (long long)(1 << i); ++t) {
        ans[j + t] = ans[j + t + (1 << i)] =
            min(ans[j + t], ans[j + t + (1 << i)]);
        ans[j + t] = min(ans[j + t], mn[j + t + (1 << i)] - mx[j + t]);
        ans[j + t + (1 << i)] =
            min(ans[j + t + (1 << i)],
                mn[j + t] - mx[j + t + (1 << i)] + (1 << i + 1));
      }
    }
    for (long long j = 0; j < (1 << k); j += (1 << i + 1)) {
      for (long long t = 0; t < (long long)(1 << i); ++t) {
        long long upmn0 = min(mn[j + t], mn[j + t + (1 << i)]);
        long long upmn1 =
            min(mn[j + t + (1 << i)] - (1 << i), mn[j + t] + (1 << i));
        long long upmx0 = max(mx[j + t], mx[j + t + (1 << i)]);
        long long upmx1 =
            max(mx[j + t + (1 << i)] - (1 << i), mx[j + t] + (1 << i));
        mn[j + t] = upmn0, mn[j + t + (1 << i)] = upmn1;
        mx[j + t] = upmx0, mx[j + t + (1 << i)] = upmx1;
      }
    }
  }
  for (long long i = 0; i < (long long)(1 << k); ++i) cout << ans[i] << " ";
  return 0;
}

### Prompt

Please formulate a cpp solution 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 long long maxn = 1 << 19;
long long n, k;
long long ans[maxn], mx[maxn], mn[maxn];
bool occ[maxn];
signed main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cin >> n >> k;
  for (long long i = 0; i < (long long)(1 << k); ++i)
    mx[i] = -1e9, mn[i] = 1e9, ans[i] = 2e9;
  for (long long i = 0; i < (long long)(n); ++i) {
    long long u;
    cin >> u;
    occ[u] = 1, mx[u] = u, mn[u] = u;
  }
  for (long long i = 0; i < k; ++i) {
    for (long long j = 0; j < (1 << k); j += (1 << i + 1)) {
      for (long long t = 0; t < (long long)(1 << i); ++t) {
        ans[j + t] = ans[j + t + (1 << i)] =
            min(ans[j + t], ans[j + t + (1 << i)]);
        ans[j + t] = min(ans[j + t], mn[j + t + (1 << i)] - mx[j + t]);
        ans[j + t + (1 << i)] =
            min(ans[j + t + (1 << i)],
                mn[j + t] - mx[j + t + (1 << i)] + (1 << i + 1));
      }
    }
    for (long long j = 0; j < (1 << k); j += (1 << i + 1)) {
      for (long long t = 0; t < (long long)(1 << i); ++t) {
        long long upmn0 = min(mn[j + t], mn[j + t + (1 << i)]);
        long long upmn1 =
            min(mn[j + t + (1 << i)] - (1 << i), mn[j + t] + (1 << i));
        long long upmx0 = max(mx[j + t], mx[j + t + (1 << i)]);
        long long upmx1 =
            max(mx[j + t + (1 << i)] - (1 << i), mx[j + t] + (1 << i));
        mn[j + t] = upmn0, mn[j + t + (1 << i)] = upmn1;
        mx[j + t] = upmx0, mx[j + t + (1 << i)] = upmx1;
      }
    }
  }
  for (long long i = 0; i < (long long)(1 << k); ++i) cout << ans[i] << " ";
  return 0;
}

```