#include <bits/stdc++.h>
using namespace std;
const int c = (1 << 20);
int n, k;
bool v[c];
void pb(vector<vector<int>> &sz, int a, int b, int c) {
  sz[0].push_back(a), sz[1].push_back(b), sz[2].push_back(c);
}
void solve(vector<vector<int>> &sz, int a) {
  sz.resize(3);
  if (a >= k) {
    if (v[a - k]) {
      pb(sz, c, a - k, a - k);
    } else {
      pb(sz, c, -1, -1);
    }
    return;
  }
  vector<vector<int>> l, r;
  solve(l, 2 * a), solve(r, 2 * a + 1);
  int si = l[0].size();
  for (int i = 0; i < si; i++) {
    int aa = min(l[0][i], r[0][i]);
    if (l[2][i] != -1 && r[1][i] != -1) {
      aa = min(aa, r[1][i] - l[2][i]);
    }
    int bb = (l[1][i] != -1 ? l[1][i] : r[1][i]),
        cc = (r[2][i] != -1 ? r[2][i] : l[2][i]);
    pb(sz, aa, bb, cc);
  }
  for (int i = 0; i < si; i++) {
    int aa = min(l[0][i], r[0][i]);
    if (r[2][i] != -1 && l[1][i] != -1) {
      aa = min(aa, l[1][i] - r[2][i] + 2 * si);
    }
    int bb = (r[1][i] != -1   ? r[1][i] - si
              : l[1][i] != -1 ? l[1][i] + si
                              : -1),
        cc = (l[2][i] != -1   ? l[2][i] + si
              : r[2][i] != -1 ? r[2][i] - si
                              : -1);
    pb(sz, aa, bb, cc);
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin >> n >> k;
  k = (1 << k);
  for (int i = 0; i < n; i++) {
    int x;
    cin >> x;
    v[x] = 1;
  }
  vector<vector<int>> ans;
  solve(ans, 1);
  for (int i = 0; i < k; i++) {
    cout << ans[0][i] << " ";
  }
  cout << "\n";
  return 0;
}

### Prompt

Develop a solution in cpp to the problem described below:

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 c = (1 << 20);
int n, k;
bool v[c];
void pb(vector<vector<int>> &sz, int a, int b, int c) {
  sz[0].push_back(a), sz[1].push_back(b), sz[2].push_back(c);
}
void solve(vector<vector<int>> &sz, int a) {
  sz.resize(3);
  if (a >= k) {
    if (v[a - k]) {
      pb(sz, c, a - k, a - k);
    } else {
      pb(sz, c, -1, -1);
    }
    return;
  }
  vector<vector<int>> l, r;
  solve(l, 2 * a), solve(r, 2 * a + 1);
  int si = l[0].size();
  for (int i = 0; i < si; i++) {
    int aa = min(l[0][i], r[0][i]);
    if (l[2][i] != -1 && r[1][i] != -1) {
      aa = min(aa, r[1][i] - l[2][i]);
    }
    int bb = (l[1][i] != -1 ? l[1][i] : r[1][i]),
        cc = (r[2][i] != -1 ? r[2][i] : l[2][i]);
    pb(sz, aa, bb, cc);
  }
  for (int i = 0; i < si; i++) {
    int aa = min(l[0][i], r[0][i]);
    if (r[2][i] != -1 && l[1][i] != -1) {
      aa = min(aa, l[1][i] - r[2][i] + 2 * si);
    }
    int bb = (r[1][i] != -1   ? r[1][i] - si
              : l[1][i] != -1 ? l[1][i] + si
                              : -1),
        cc = (l[2][i] != -1   ? l[2][i] + si
              : r[2][i] != -1 ? r[2][i] - si
                              : -1);
    pb(sz, aa, bb, cc);
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin >> n >> k;
  k = (1 << k);
  for (int i = 0; i < n; i++) {
    int x;
    cin >> x;
    v[x] = 1;
  }
  vector<vector<int>> ans;
  solve(ans, 1);
  for (int i = 0; i < k; i++) {
    cout << ans[0][i] << " ";
  }
  cout << "\n";
  return 0;
}

```