#include <bits/stdc++.h>
using namespace std;
const int MAXN = 2000005;
const int MAXK = 20;
const int INF = 0x3f3f3f3f;
int inline read() {
  int x = 0, f = 0;
  char ch = getchar();
  while (ch < '0' || ch > '9') f ^= (ch == '-'), ch = getchar();
  while (ch >= '0' && ch <= '9') x = x * 10 + (ch ^ 48), ch = getchar();
  return f ? -x : x;
}
int n, k, a[MAXN];
int f[MAXN], minn[MAXN], maxn[MAXN];
signed main() {
  memset(f, 0x3f, sizeof(f));
  memset(minn, 0x3f, sizeof(minn));
  memset(maxn, ~0x3f, sizeof(maxn));
  n = read();
  k = read();
  for (int i = 1; i <= n; ++i) {
    int x = read();
    maxn[x] = minn[x] = 0;
  }
  for (int i = 0; i < k; ++i)
    for (int x = 0; x < (1 << k); ++x)
      if (x & (1 << i)) {
        ;
        int y = x ^ (1 << i);
        f[x] = f[y] = min(f[x], f[y]);
        f[x] = min(f[x], minn[y] - maxn[x] + (1 << i));
        f[y] = min(f[y], minn[x] - maxn[y] + (1 << i));
        int mnx = minn[x], mny = minn[y], mxx = maxn[x], mxy = maxn[y];
        minn[x] = min(mnx, mny + (1 << i));
        minn[y] = min(mny, mnx + (1 << i));
        maxn[x] = max(mxx, mxy + (1 << i));
        maxn[y] = max(mxy, mxx + (1 << i));
      }
  for (int i = 0; i < (1 << k); ++i) printf("%d ", f[i]);
}

### 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;
const int MAXN = 2000005;
const int MAXK = 20;
const int INF = 0x3f3f3f3f;
int inline read() {
  int x = 0, f = 0;
  char ch = getchar();
  while (ch < '0' || ch > '9') f ^= (ch == '-'), ch = getchar();
  while (ch >= '0' && ch <= '9') x = x * 10 + (ch ^ 48), ch = getchar();
  return f ? -x : x;
}
int n, k, a[MAXN];
int f[MAXN], minn[MAXN], maxn[MAXN];
signed main() {
  memset(f, 0x3f, sizeof(f));
  memset(minn, 0x3f, sizeof(minn));
  memset(maxn, ~0x3f, sizeof(maxn));
  n = read();
  k = read();
  for (int i = 1; i <= n; ++i) {
    int x = read();
    maxn[x] = minn[x] = 0;
  }
  for (int i = 0; i < k; ++i)
    for (int x = 0; x < (1 << k); ++x)
      if (x & (1 << i)) {
        ;
        int y = x ^ (1 << i);
        f[x] = f[y] = min(f[x], f[y]);
        f[x] = min(f[x], minn[y] - maxn[x] + (1 << i));
        f[y] = min(f[y], minn[x] - maxn[y] + (1 << i));
        int mnx = minn[x], mny = minn[y], mxx = maxn[x], mxy = maxn[y];
        minn[x] = min(mnx, mny + (1 << i));
        minn[y] = min(mny, mnx + (1 << i));
        maxn[x] = max(mxx, mxy + (1 << i));
        maxn[y] = max(mxy, mxx + (1 << i));
      }
  for (int i = 0; i < (1 << k); ++i) printf("%d ", f[i]);
}

```