#include <bits/stdc++.h>
using namespace std;
const long long mod = 998244353;
const int iu = 1e6 + 1;
const int N = 2e6 + 1;
int n, z;
int a[N];
int mn[N], mx[N];
int ans[N];
vector<int> f[N];
int mwah[N];
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cin >> n >> z;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
  }
  int s = 0;
  for (int i = 0; i < (1 << z); i++) ans[i] = 1e9;
  while ((1 << (s + 1)) < n) s++;
  for (int i = 0; i < (1 << (z - s)); i++) {
    for (int j = 0; j < (1 << s); j++) {
      f[j].clear();
    }
    for (int j = 1; j <= n; j++) {
      int cur = (a[j] ^ i);
      f[cur >> (z - s)].push_back(cur % (1 << (z - s)));
    }
    int funny = 1e9;
    for (int j = 0; j < (1 << s); j++) {
      sort(f[j].begin(), f[j].end());
      for (int k = 1; k < f[j].size(); k++) {
        funny = min(funny, f[j][k] - f[j][k - 1]);
      }
    }
    for (int k = 0; k < (1 << s); k++) {
      int cur = (k << (z - s)) + i;
      ans[cur] = min(ans[cur], funny);
    }
    for (int j = 0; j < s; j++) {
      for (int k = 0; k < (1 << s); k++) {
        mwah[k] = 1e9;
      }
      for (int k = 0; k < (1 << s); k++) {
        int x1 = k;
        int x2 = k ^ ((1 << (j + 1)) - 1);
        if (f[x1].empty() || f[x2].empty()) continue;
        int bad = (((1 << j) - 1) ^ k) % (1 << (j + 1));
        mwah[bad] = min(mwah[bad], f[x2][0] - f[x1].back() + (1 << (z - s)));
      }
      for (int k = 0; k < (1 << s); k++) {
        int cur = (k << (z - s)) + i;
        ans[cur] = min(ans[cur], mwah[k % (1 << (j + 1))]);
      }
    }
  }
  for (int i = 0; i < (1 << z); i++) cout << ans[i] << ' ';
  cout << '\n';
}

### 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 long long mod = 998244353;
const int iu = 1e6 + 1;
const int N = 2e6 + 1;
int n, z;
int a[N];
int mn[N], mx[N];
int ans[N];
vector<int> f[N];
int mwah[N];
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cin >> n >> z;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
  }
  int s = 0;
  for (int i = 0; i < (1 << z); i++) ans[i] = 1e9;
  while ((1 << (s + 1)) < n) s++;
  for (int i = 0; i < (1 << (z - s)); i++) {
    for (int j = 0; j < (1 << s); j++) {
      f[j].clear();
    }
    for (int j = 1; j <= n; j++) {
      int cur = (a[j] ^ i);
      f[cur >> (z - s)].push_back(cur % (1 << (z - s)));
    }
    int funny = 1e9;
    for (int j = 0; j < (1 << s); j++) {
      sort(f[j].begin(), f[j].end());
      for (int k = 1; k < f[j].size(); k++) {
        funny = min(funny, f[j][k] - f[j][k - 1]);
      }
    }
    for (int k = 0; k < (1 << s); k++) {
      int cur = (k << (z - s)) + i;
      ans[cur] = min(ans[cur], funny);
    }
    for (int j = 0; j < s; j++) {
      for (int k = 0; k < (1 << s); k++) {
        mwah[k] = 1e9;
      }
      for (int k = 0; k < (1 << s); k++) {
        int x1 = k;
        int x2 = k ^ ((1 << (j + 1)) - 1);
        if (f[x1].empty() || f[x2].empty()) continue;
        int bad = (((1 << j) - 1) ^ k) % (1 << (j + 1));
        mwah[bad] = min(mwah[bad], f[x2][0] - f[x1].back() + (1 << (z - s)));
      }
      for (int k = 0; k < (1 << s); k++) {
        int cur = (k << (z - s)) + i;
        ans[cur] = min(ans[cur], mwah[k % (1 << (j + 1))]);
      }
    }
  }
  for (int i = 0; i < (1 << z); i++) cout << ans[i] << ' ';
  cout << '\n';
}

```