#include <bits/stdc++.h>
using namespace std;
int cnt[(1 << 19)], mn[(1 << 19)], mx[(1 << 19)], ans[(1 << 19)];
void solve(int L, int R) {
  if (L + 1 == R) {
    ans[L] = 0x3f3f3f3f;
    mn[L] = cnt[L] ? 0 : 0x3f3f3f3f;
    mx[L] = cnt[L] ? 0 : -0x3f3f3f3f;
  } else {
    int M = (L + R) / 2, i, H = M - L;
    solve(L, M), solve(M, R);
    for (i = L; i < M; ++i) {
      int mn1 = mn[i], mx1 = mx[i], ans1 = ans[i], mn2 = mn[i + H],
          mx2 = mx[i + H], ans2 = ans[i + H];
      mn[i] = min(mn1, mn2 + H);
      mx[i] = max(mx1, mx2 + H);
      ans[i] = min({ans1, ans2, mn2 + H - mx1});
      mn[i + H] = min(mn2, mn1 + H);
      mx[i + H] = max(mx2, mx1 + H);
      ans[i + H] = min({ans1, ans2, mn1 + H - mx2});
    }
  }
}
int main() {
  int n, k, i;
  scanf("%d%d", &n, &k);
  while (n--) {
    scanf("%d", &i);
    ++cnt[i];
  }
  solve(0, 1 << k);
  for (i = 0; i < (1 << k); ++i) printf("%d ", ans[i]);
  printf("\n");
  return 0;
}

### 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 cnt[(1 << 19)], mn[(1 << 19)], mx[(1 << 19)], ans[(1 << 19)];
void solve(int L, int R) {
  if (L + 1 == R) {
    ans[L] = 0x3f3f3f3f;
    mn[L] = cnt[L] ? 0 : 0x3f3f3f3f;
    mx[L] = cnt[L] ? 0 : -0x3f3f3f3f;
  } else {
    int M = (L + R) / 2, i, H = M - L;
    solve(L, M), solve(M, R);
    for (i = L; i < M; ++i) {
      int mn1 = mn[i], mx1 = mx[i], ans1 = ans[i], mn2 = mn[i + H],
          mx2 = mx[i + H], ans2 = ans[i + H];
      mn[i] = min(mn1, mn2 + H);
      mx[i] = max(mx1, mx2 + H);
      ans[i] = min({ans1, ans2, mn2 + H - mx1});
      mn[i + H] = min(mn2, mn1 + H);
      mx[i + H] = max(mx2, mx1 + H);
      ans[i + H] = min({ans1, ans2, mn1 + H - mx2});
    }
  }
}
int main() {
  int n, k, i;
  scanf("%d%d", &n, &k);
  while (n--) {
    scanf("%d", &i);
    ++cnt[i];
  }
  solve(0, 1 << k);
  for (i = 0; i < (1 << k); ++i) printf("%d ", ans[i]);
  printf("\n");
  return 0;
}

```