#include <bits/stdc++.h>
using namespace std;
int n, m, a[1000005], tans[22][1000005], b[1000005], c[1000005];
void solve(int k, int l, int r) {
  if (l >= r) return;
  int lim = (1 << (k - 1));
  for (int i = 0; i < (1 << k); i++) b[i] = -1e9, c[i] = 1e9;
  int mid = l - 1;
  while (mid < r && ((a[mid + 1] >> (k - 1)) & 1) == 0) mid++;
  int mn = max(0, lim - m), mx = min((1 << k) - 1, lim + m);
  for (int i = l; i <= r; i++) {
    int u = a[i] & ((1 << k) - 1);
    for (int j = mn; j < lim; j++) b[u ^ j] = max(b[u ^ j], j);
    for (int j = lim; j <= mx; j++) c[u ^ j] = min(c[u ^ j], j);
  }
  for (int i = 0; i < (1 << k); i++) tans[k][i] = min(tans[k][i], c[i] - b[i]);
  solve(k - 1, l, mid);
  solve(k - 1, mid + 1, r);
}
int k;
int main() {
  scanf("%d%d", &n, &k);
  m = ((1 << k) - 1) / (n - 1);
  for (int j = 1; j <= k; j++)
    for (int i = 0; i < (1 << j); i++) tans[j][i] = m;
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  sort(a + 1, a + n + 1);
  solve(k, 1, n);
  for (int i = 0; i < (1 << k); i++) {
    int ans = 1e9;
    for (int j = 1; j <= k; j++) ans = min(ans, tans[j][i & ((1 << j) - 1)]);
    printf("%d ", ans);
  }
  return 0;
}

### Prompt

Please provide a CPP coded solution 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 namespace std;
int n, m, a[1000005], tans[22][1000005], b[1000005], c[1000005];
void solve(int k, int l, int r) {
  if (l >= r) return;
  int lim = (1 << (k - 1));
  for (int i = 0; i < (1 << k); i++) b[i] = -1e9, c[i] = 1e9;
  int mid = l - 1;
  while (mid < r && ((a[mid + 1] >> (k - 1)) & 1) == 0) mid++;
  int mn = max(0, lim - m), mx = min((1 << k) - 1, lim + m);
  for (int i = l; i <= r; i++) {
    int u = a[i] & ((1 << k) - 1);
    for (int j = mn; j < lim; j++) b[u ^ j] = max(b[u ^ j], j);
    for (int j = lim; j <= mx; j++) c[u ^ j] = min(c[u ^ j], j);
  }
  for (int i = 0; i < (1 << k); i++) tans[k][i] = min(tans[k][i], c[i] - b[i]);
  solve(k - 1, l, mid);
  solve(k - 1, mid + 1, r);
}
int k;
int main() {
  scanf("%d%d", &n, &k);
  m = ((1 << k) - 1) / (n - 1);
  for (int j = 1; j <= k; j++)
    for (int i = 0; i < (1 << j); i++) tans[j][i] = m;
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  sort(a + 1, a + n + 1);
  solve(k, 1, n);
  for (int i = 0; i < (1 << k); i++) {
    int ans = 1e9;
    for (int j = 1; j <= k; j++) ans = min(ans, tans[j][i & ((1 << j) - 1)]);
    printf("%d ", ans);
  }
  return 0;
}

```