#include <bits/stdc++.h>
using namespace std;
int k;
struct node {
  node* c[2];
};
node* trie;
void ins(int r) {
  node* f = trie;
  for (int j = k - 1; j >= 0; j--) {
    if (!f->c[1 & (r >> j)]) {
      f->c[1 & (r >> j)] = new node();
      f->c[1 & (r >> j)]->c[0] = f->c[1 & (r >> j)]->c[1] = nullptr;
    }
    f = f->c[1 & (r >> j)];
  }
}
vector<vector<int> > ans;
void solve(node* l, node* r, int st, int x, int d, int diff) {
  if (d == k) {
    ans[st][x] = min(ans[st][x], diff);
  } else {
    for (int j = 0; j < 2; j++) {
      if (l->c[1 ^ j] && r->c[0 ^ j]) {
        solve(l->c[1 ^ j], r->c[0 ^ j], st, x ^ (j << (k - 1 - d)), d + 1,
              diff - (1 << (k - 1 - d)));
        continue;
      }
      if (l->c[1 ^ j] && r->c[1 ^ j]) {
        solve(l->c[1 ^ j], r->c[1 ^ j], st, x ^ (j << (k - 1 - d)), d + 1,
              diff);
        continue;
      }
      if (l->c[0 ^ j] && r->c[0 ^ j]) {
        solve(l->c[0 ^ j], r->c[0 ^ j], st, x ^ (j << (k - 1 - d)), d + 1,
              diff);
        continue;
      }
      if (l->c[0 ^ j] && r->c[1 ^ j]) {
        solve(l->c[0 ^ j], r->c[1 ^ j], st, x ^ (j << (k - 1 - d)), d + 1,
              diff + (1 << (k - 1 - d)));
        continue;
      }
    }
  }
}
void tr(node* v, int r, int d) {
  if (v->c[0] && v->c[1]) {
    solve(v->c[0], v->c[1], d, 0, d + 1, (1 << (k - 1 - d)));
    solve(v->c[1], v->c[0], d, (1 << (k - 1 - d)), d + 1, (1 << (k - 1 - d)));
  }
  if (v->c[0]) tr(v->c[0], 2 * r + 0, d + 1);
  if (v->c[1]) tr(v->c[1], 2 * r + 1, d + 1);
}
int main() {
  ios_base::sync_with_stdio(false), cin.tie(nullptr);
  trie = new node();
  trie->c[0] = trie->c[1] = nullptr;
  int n;
  cin >> n >> k;
  ans.assign(k, vector<int>(1 << k, int(1e8)));
  vector<int> a(n);
  for (int& x : a) cin >> x;
  sort(a.begin(), a.end());
  for (int x : a) ins(x);
  assert(trie != nullptr);
  tr(trie, 0, 0);
  for (int i = 0; i < (1 << k); i++) {
    int best = int(1e8);
    for (int j = 0; j < k; j++) {
      best = min(best, ans[j][i % (1 << (k - j))]);
    }
    cout << best << ' ';
  }
  cout << '\n';
}

### Prompt

In Cpp, your task is to solve 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 k;
struct node {
  node* c[2];
};
node* trie;
void ins(int r) {
  node* f = trie;
  for (int j = k - 1; j >= 0; j--) {
    if (!f->c[1 & (r >> j)]) {
      f->c[1 & (r >> j)] = new node();
      f->c[1 & (r >> j)]->c[0] = f->c[1 & (r >> j)]->c[1] = nullptr;
    }
    f = f->c[1 & (r >> j)];
  }
}
vector<vector<int> > ans;
void solve(node* l, node* r, int st, int x, int d, int diff) {
  if (d == k) {
    ans[st][x] = min(ans[st][x], diff);
  } else {
    for (int j = 0; j < 2; j++) {
      if (l->c[1 ^ j] && r->c[0 ^ j]) {
        solve(l->c[1 ^ j], r->c[0 ^ j], st, x ^ (j << (k - 1 - d)), d + 1,
              diff - (1 << (k - 1 - d)));
        continue;
      }
      if (l->c[1 ^ j] && r->c[1 ^ j]) {
        solve(l->c[1 ^ j], r->c[1 ^ j], st, x ^ (j << (k - 1 - d)), d + 1,
              diff);
        continue;
      }
      if (l->c[0 ^ j] && r->c[0 ^ j]) {
        solve(l->c[0 ^ j], r->c[0 ^ j], st, x ^ (j << (k - 1 - d)), d + 1,
              diff);
        continue;
      }
      if (l->c[0 ^ j] && r->c[1 ^ j]) {
        solve(l->c[0 ^ j], r->c[1 ^ j], st, x ^ (j << (k - 1 - d)), d + 1,
              diff + (1 << (k - 1 - d)));
        continue;
      }
    }
  }
}
void tr(node* v, int r, int d) {
  if (v->c[0] && v->c[1]) {
    solve(v->c[0], v->c[1], d, 0, d + 1, (1 << (k - 1 - d)));
    solve(v->c[1], v->c[0], d, (1 << (k - 1 - d)), d + 1, (1 << (k - 1 - d)));
  }
  if (v->c[0]) tr(v->c[0], 2 * r + 0, d + 1);
  if (v->c[1]) tr(v->c[1], 2 * r + 1, d + 1);
}
int main() {
  ios_base::sync_with_stdio(false), cin.tie(nullptr);
  trie = new node();
  trie->c[0] = trie->c[1] = nullptr;
  int n;
  cin >> n >> k;
  ans.assign(k, vector<int>(1 << k, int(1e8)));
  vector<int> a(n);
  for (int& x : a) cin >> x;
  sort(a.begin(), a.end());
  for (int x : a) ins(x);
  assert(trie != nullptr);
  tr(trie, 0, 0);
  for (int i = 0; i < (1 << k); i++) {
    int best = int(1e8);
    for (int j = 0; j < k; j++) {
      best = min(best, ans[j][i % (1 << (k - j))]);
    }
    cout << best << ' ';
  }
  cout << '\n';
}

```