#include <bits/stdc++.h>
using ll = long long;
int const nmax = 19;
int v[1 << nmax];
struct Info {
  int smin;
  int pref;
  int suff;
} dp[1 << nmax];
int main() {
  int n, k;
  std::cin >> n >> k;
  for (int i = 1; i <= n; i++) {
    int val;
    std::cin >> val;
    v[val] = 1;
  }
  for (int i = 0; i < (1 << k); i++)
    if (v[i] == 1) {
      dp[i] = {(1 << k), 0, 0};
    } else {
      dp[i] = {(1 << k), (1 << k), (1 << k)};
    }
  for (int jump = 2; jump <= (1 << k); jump *= 2) {
    int halfjump = jump / 2;
    for (int i = 0; i < (1 << k); i += jump) {
      for (int j = i; j < i + halfjump; j++) {
        Info a = dp[j], b = dp[j + halfjump];
        dp[j].smin = std::min(std::min(a.smin, b.smin), a.suff + b.pref + 1);
        dp[j].pref = std::min(a.pref, halfjump + b.pref);
        dp[j].suff = std::min(a.suff + halfjump, b.suff);
        dp[j + halfjump].smin =
            std::min(std::min(a.smin, b.smin), a.pref + b.suff + 1);
        dp[j + halfjump].pref = std::min(b.pref, halfjump + a.pref);
        dp[j + halfjump].suff = std::min(a.suff, halfjump + b.suff);
      }
    }
  }
  for (int i = 0; i < (1 << k); i++) std::cout << dp[i].smin << " ";
  std::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 ll = long long;
int const nmax = 19;
int v[1 << nmax];
struct Info {
  int smin;
  int pref;
  int suff;
} dp[1 << nmax];
int main() {
  int n, k;
  std::cin >> n >> k;
  for (int i = 1; i <= n; i++) {
    int val;
    std::cin >> val;
    v[val] = 1;
  }
  for (int i = 0; i < (1 << k); i++)
    if (v[i] == 1) {
      dp[i] = {(1 << k), 0, 0};
    } else {
      dp[i] = {(1 << k), (1 << k), (1 << k)};
    }
  for (int jump = 2; jump <= (1 << k); jump *= 2) {
    int halfjump = jump / 2;
    for (int i = 0; i < (1 << k); i += jump) {
      for (int j = i; j < i + halfjump; j++) {
        Info a = dp[j], b = dp[j + halfjump];
        dp[j].smin = std::min(std::min(a.smin, b.smin), a.suff + b.pref + 1);
        dp[j].pref = std::min(a.pref, halfjump + b.pref);
        dp[j].suff = std::min(a.suff + halfjump, b.suff);
        dp[j + halfjump].smin =
            std::min(std::min(a.smin, b.smin), a.pref + b.suff + 1);
        dp[j + halfjump].pref = std::min(b.pref, halfjump + a.pref);
        dp[j + halfjump].suff = std::min(a.suff, halfjump + b.suff);
      }
    }
  }
  for (int i = 0; i < (1 << k); i++) std::cout << dp[i].smin << " ";
  std::cout << '\n';
  return 0;
}

```