#include <bits/stdc++.h>
using namespace std;
template <class T>
bool umax(T &x, T y) {
  return y > x ? x = y, 1 : 0;
}
template <class T>
bool umin(T &x, T y) {
  return y < x ? x = y, 1 : 0;
}
const int N = 1 << 19;
const int inf = 1e9;
int n, k, mx[N], mi[N], f[N];
int main() {
  ios::sync_with_stdio(false);
  cin >> n >> k;
  for (int i = 0; i < (1 << k); ++i) {
    mx[i] = -inf;
    mi[i] = inf;
    f[i] = inf;
  }
  while (n--) {
    int x;
    cin >> x;
    mx[x] = mi[x] = 0;
  }
  for (int i = 0; i < k; ++i) {
    for (int y = 0; y < (1 << k); ++y)
      if (y >> i & 1) {
        int x = y ^ (1 << i);
        f[x] = f[y] = min(f[x], f[y]);
        umin(f[x], mi[y] + (1 << i) - mx[x]);
        umin(f[y], mi[x] + (1 << i) - mx[y]);
        int mix = mi[x], miy = mi[y], mxx = mx[x], mxy = mx[y];
        umin(mi[x], miy + (1 << i));
        umin(mi[y], mix + (1 << i));
        umax(mx[x], mxy + (1 << i));
        umax(mx[y], mxx + (1 << i));
      }
  }
  for (int i = 0; i < (1 << k); ++i) cout << f[i] << " ";
  cout << "\n";
  return 0;
}

### 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;
template <class T>
bool umax(T &x, T y) {
  return y > x ? x = y, 1 : 0;
}
template <class T>
bool umin(T &x, T y) {
  return y < x ? x = y, 1 : 0;
}
const int N = 1 << 19;
const int inf = 1e9;
int n, k, mx[N], mi[N], f[N];
int main() {
  ios::sync_with_stdio(false);
  cin >> n >> k;
  for (int i = 0; i < (1 << k); ++i) {
    mx[i] = -inf;
    mi[i] = inf;
    f[i] = inf;
  }
  while (n--) {
    int x;
    cin >> x;
    mx[x] = mi[x] = 0;
  }
  for (int i = 0; i < k; ++i) {
    for (int y = 0; y < (1 << k); ++y)
      if (y >> i & 1) {
        int x = y ^ (1 << i);
        f[x] = f[y] = min(f[x], f[y]);
        umin(f[x], mi[y] + (1 << i) - mx[x]);
        umin(f[y], mi[x] + (1 << i) - mx[y]);
        int mix = mi[x], miy = mi[y], mxx = mx[x], mxy = mx[y];
        umin(mi[x], miy + (1 << i));
        umin(mi[y], mix + (1 << i));
        umax(mx[x], mxy + (1 << i));
        umax(mx[y], mxx + (1 << i));
      }
  }
  for (int i = 0; i < (1 << k); ++i) cout << f[i] << " ";
  cout << "\n";
  return 0;
}

```