#include <bits/stdc++.h>
using namespace std;
int store[20][1 << 20];
struct trie {
  int k;
  struct Node {
    Node* ch[2];
    Node() { ch[0] = ch[1] = NULL; }
  };
  Node* root;
  trie(int k) : k(k) { root = new Node(); }
  void insert(int x) {
    Node* cur = root;
    for (int i = k - 1; i >= 0; i--) {
      int side = x >> i & 1;
      if (!cur->ch[side]) cur->ch[side] = new Node();
      cur = cur->ch[side];
    }
  }
  void eval() {
    vector<int> ret(1 << k);
    for (int i = 0; i <= k; i++) fill(store[i], store[i] + (1 << k), 1 << 30);
    function<vector<pair<int, int> >(Node*, int)> dfs = [&](Node* node, int h) {
      if (!node) return vector<pair<int, int> >();
      if (h == 0) {
        vector<pair<int, int> > ret = {{0, 0}};
        return ret;
      }
      vector<pair<int, int> > A = dfs(node->ch[0], h - 1),
                              B = dfs(node->ch[1], h - 1);
      vector<pair<int, int> > ret(1 << h);
      for (int x = 0; x < (1 << h); x++) {
        int add = 1 << (h - 1);
        int rem = x & (add - 1);
        if (x >> (h - 1) & 1) {
          ret[x].first = node->ch[1] ? B[rem].first : A[rem].first + add;
          ret[x].second = node->ch[0] ? A[rem].second + add : B[rem].second;
        } else {
          ret[x].first = node->ch[0] ? A[rem].first : B[rem].first + add;
          ret[x].second = node->ch[1] ? B[rem].second + add : A[rem].second;
        }
        if (!A.empty() && !B.empty()) {
          if (x >> (h - 1) & 1)
            store[h][x] = min(store[h][x], A[rem].first + add - B[rem].second);
          else
            store[h][x] = min(store[h][x], B[rem].first + add - A[rem].second);
        }
        assert(max(ret[x].first, ret[x].second) < (1 << h) &&
               min(ret[x].first, ret[x].second) >= 0);
      }
      return ret;
    };
    dfs(root, k);
  }
};
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int n, k;
  cin >> n >> k;
  trie T(k);
  for (int i = 0; i < n; i++) {
    int x;
    cin >> x;
    T.insert(x);
  }
  T.eval();
  for (int x = 0; x < (1 << k); x++) {
    int ans = 1 << 30;
    for (int h = k; h; h--) {
      ans = min(ans, store[h][x & ((1 << h) - 1)]);
    }
    cout << ans << " ";
  }
  cout << '\n';
}

### 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;
int store[20][1 << 20];
struct trie {
  int k;
  struct Node {
    Node* ch[2];
    Node() { ch[0] = ch[1] = NULL; }
  };
  Node* root;
  trie(int k) : k(k) { root = new Node(); }
  void insert(int x) {
    Node* cur = root;
    for (int i = k - 1; i >= 0; i--) {
      int side = x >> i & 1;
      if (!cur->ch[side]) cur->ch[side] = new Node();
      cur = cur->ch[side];
    }
  }
  void eval() {
    vector<int> ret(1 << k);
    for (int i = 0; i <= k; i++) fill(store[i], store[i] + (1 << k), 1 << 30);
    function<vector<pair<int, int> >(Node*, int)> dfs = [&](Node* node, int h) {
      if (!node) return vector<pair<int, int> >();
      if (h == 0) {
        vector<pair<int, int> > ret = {{0, 0}};
        return ret;
      }
      vector<pair<int, int> > A = dfs(node->ch[0], h - 1),
                              B = dfs(node->ch[1], h - 1);
      vector<pair<int, int> > ret(1 << h);
      for (int x = 0; x < (1 << h); x++) {
        int add = 1 << (h - 1);
        int rem = x & (add - 1);
        if (x >> (h - 1) & 1) {
          ret[x].first = node->ch[1] ? B[rem].first : A[rem].first + add;
          ret[x].second = node->ch[0] ? A[rem].second + add : B[rem].second;
        } else {
          ret[x].first = node->ch[0] ? A[rem].first : B[rem].first + add;
          ret[x].second = node->ch[1] ? B[rem].second + add : A[rem].second;
        }
        if (!A.empty() && !B.empty()) {
          if (x >> (h - 1) & 1)
            store[h][x] = min(store[h][x], A[rem].first + add - B[rem].second);
          else
            store[h][x] = min(store[h][x], B[rem].first + add - A[rem].second);
        }
        assert(max(ret[x].first, ret[x].second) < (1 << h) &&
               min(ret[x].first, ret[x].second) >= 0);
      }
      return ret;
    };
    dfs(root, k);
  }
};
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int n, k;
  cin >> n >> k;
  trie T(k);
  for (int i = 0; i < n; i++) {
    int x;
    cin >> x;
    T.insert(x);
  }
  T.eval();
  for (int x = 0; x < (1 << k); x++) {
    int ans = 1 << 30;
    for (int h = k; h; h--) {
      ans = min(ans, store[h][x & ((1 << h) - 1)]);
    }
    cout << ans << " ";
  }
  cout << '\n';
}

```