#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using ld = long double;
const int INF = 1e9;
struct Node {
  int mn = INF;
  int mx = -INF;
  int minAdj = INF;
  Node() {}
  Node(const Node& l, const Node& r, int bit) {
    mn = min(l.mn, r.mn + bit);
    mx = max(l.mx, r.mx + bit);
    minAdj = min(l.minAdj, r.minAdj);
    minAdj = min(minAdj, r.mn + bit - l.mx);
  }
};
vector<bool> has;
vector<Node> solve(int l, int a) {
  if (l < 0) {
    vector<Node> ret(1);
    if (has[a]) ret[0].mn = ret[0].mx = 0;
    return ret;
  }
  auto L = solve(l - 1, a * 2);
  auto R = solve(l - 1, a * 2 + 1);
  vector<Node> ret(L.size() * 2);
  for (int i = 0; i < L.size(); ++i) {
    ret[i] = Node(L[i], R[i], 1 << l);
    ret[i + L.size()] = Node(R[i], L[i], 1 << l);
  }
  return ret;
}
int main() {
  int n, k;
  scanf("%d%d", &n, &k);
  has.assign(1 << k, false);
  for (int i = 0; i < n; ++i) {
    int a;
    scanf("%d", &a);
    has[a] = true;
  }
  auto ans = solve(k - 1, 0);
  for (Node& a : ans) printf("%d ", a.minAdj);
}

### Prompt

Your task is to create 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;
using ll = long long;
using ld = long double;
const int INF = 1e9;
struct Node {
  int mn = INF;
  int mx = -INF;
  int minAdj = INF;
  Node() {}
  Node(const Node& l, const Node& r, int bit) {
    mn = min(l.mn, r.mn + bit);
    mx = max(l.mx, r.mx + bit);
    minAdj = min(l.minAdj, r.minAdj);
    minAdj = min(minAdj, r.mn + bit - l.mx);
  }
};
vector<bool> has;
vector<Node> solve(int l, int a) {
  if (l < 0) {
    vector<Node> ret(1);
    if (has[a]) ret[0].mn = ret[0].mx = 0;
    return ret;
  }
  auto L = solve(l - 1, a * 2);
  auto R = solve(l - 1, a * 2 + 1);
  vector<Node> ret(L.size() * 2);
  for (int i = 0; i < L.size(); ++i) {
    ret[i] = Node(L[i], R[i], 1 << l);
    ret[i + L.size()] = Node(R[i], L[i], 1 << l);
  }
  return ret;
}
int main() {
  int n, k;
  scanf("%d%d", &n, &k);
  has.assign(1 << k, false);
  for (int i = 0; i < n; ++i) {
    int a;
    scanf("%d", &a);
    has[a] = true;
  }
  auto ans = solve(k - 1, 0);
  for (Node& a : ans) printf("%d ", a.minAdj);
}

```