#include <bits/stdc++.h>
using namespace std;
const int MK = 20, MN = 1 << MK, INF = 0x3f3f3f3f;
int N, K, M;
struct node {
  int mn, mx, ans;
  node(int mn = -1, int mx = -1, int ans = INF) : mn(mn), mx(mx), ans(ans) {}
  operator bool() { return mn > -1; }
} tree[MK][MN];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> N >> K;
  M = 1 << K;
  for (int i = 0, x; i < N; i++) {
    cin >> x;
    tree[0][x] = {x, x, INF};
  }
  for (int lv = 1, sz, psz; sz = 1 << lv, psz = sz >> 1, lv <= K; lv++) {
    for (int i = 0; i < M; i += sz) {
      for (int bit = 0; bit < psz; ++bit) {
        node a = tree[lv - 1][i | bit], b = tree[lv - 1][i | psz | bit],
             &cur0 = tree[lv][i | bit], &cur1 = tree[lv][i | psz | bit];
        if (a && b) {
          cur0 = {a.mn, b.mx, min(b.mn - a.mx, min(a.ans, b.ans))};
          cur1 = {b.mn ^ psz, a.mx ^ psz,
                  min(sz - b.mx + a.mn, min(a.ans, b.ans))};
        } else {
          cur0 = a ? a : b;
          cur1 = cur0 ? node(cur0.mn ^ psz, cur0.mx ^ psz, cur0.ans) : cur0;
        }
      }
    }
  }
  for (int bit = 0; bit < M; ++bit) {
    printf("%d%c", tree[K][bit].ans, " \n"[bit == M - 1]);
  }
  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 MK = 20, MN = 1 << MK, INF = 0x3f3f3f3f;
int N, K, M;
struct node {
  int mn, mx, ans;
  node(int mn = -1, int mx = -1, int ans = INF) : mn(mn), mx(mx), ans(ans) {}
  operator bool() { return mn > -1; }
} tree[MK][MN];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> N >> K;
  M = 1 << K;
  for (int i = 0, x; i < N; i++) {
    cin >> x;
    tree[0][x] = {x, x, INF};
  }
  for (int lv = 1, sz, psz; sz = 1 << lv, psz = sz >> 1, lv <= K; lv++) {
    for (int i = 0; i < M; i += sz) {
      for (int bit = 0; bit < psz; ++bit) {
        node a = tree[lv - 1][i | bit], b = tree[lv - 1][i | psz | bit],
             &cur0 = tree[lv][i | bit], &cur1 = tree[lv][i | psz | bit];
        if (a && b) {
          cur0 = {a.mn, b.mx, min(b.mn - a.mx, min(a.ans, b.ans))};
          cur1 = {b.mn ^ psz, a.mx ^ psz,
                  min(sz - b.mx + a.mn, min(a.ans, b.ans))};
        } else {
          cur0 = a ? a : b;
          cur1 = cur0 ? node(cur0.mn ^ psz, cur0.mx ^ psz, cur0.ans) : cur0;
        }
      }
    }
  }
  for (int bit = 0; bit < M; ++bit) {
    printf("%d%c", tree[K][bit].ans, " \n"[bit == M - 1]);
  }
  return 0;
}

```