#include <bits/stdc++.h>
using std::min;
using std::swap;
int s[11111111][2], dl[11111111], dr[11111111], minv[11111111];
int sz[11111111];
int k, n;
int rv[1 << 21];
int a[1 << 21];
int cp = 1;
void pushu(int p) {
  if (s[p][0] && s[p][1])
    dl[p] = dl[s[p][0]], dr[p] = dr[s[p][1]],
    minv[p] =
        min(dr[s[p][0]] + dl[s[p][1]] + 1, min(minv[s[p][0]], minv[s[p][1]]));
  else if (s[p][0])
    dl[p] = dl[s[p][0]], dr[p] = dr[s[p][0]] + sz[p], minv[p] = minv[s[p][0]];
  else if (s[p][1])
    dr[p] = dr[s[p][1]], dl[p] = dl[s[p][1]] + sz[p], minv[p] = minv[s[p][1]];
}
void ins(int x, int p = 1, int i = k - 1) {
  if (i < 0) {
    dl[p] = dr[p] = 0, minv[p] = 1e9;
    return;
  }
  sz[p] = 1 << i;
  int d = (x >> i) & 1;
  if (!s[p][d]) s[p][d] = ++cp;
  ins(x - (d << i), s[p][d], i - 1), pushu(p);
}
void init() {
  for (register int i = 1; i <= n; i++) ins(a[i]);
}
void flip(int dep, int p = 1, int i = k - 1) {
  if (!p) return;
  if (i == dep) {
    swap(s[p][0], s[p][1]), pushu(p);
    return;
  }
  flip(dep, s[p][0], i - 1), flip(dep, s[p][1], i - 1), pushu(p);
}
int T;
int recalc(int S) {
  for (register int i = 0; i < k; i++)
    if (((S ^ T) >> i) & 1) flip(i);
  T = S;
  return minv[1];
}
int ans[1 << 21];
int main() {
  scanf("%d%d", &n, &k);
  for (register int i = 1; i <= n; i++) scanf("%d", a + i);
  init();
  for (register int i = 1; i < (1 << k); i++)
    rv[i] = (rv[i >> 1] >> 1) | ((i & 1) ? (1 << (k - 1)) : 0);
  for (register int i = 0; i < (1 << k); i++) ans[rv[i]] = recalc(rv[i]);
  for (register int i = 0; i < (1 << k); i++) printf("%d ", ans[i]);
  puts("");
}

### Prompt

Your challenge is to write 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 std::min;
using std::swap;
int s[11111111][2], dl[11111111], dr[11111111], minv[11111111];
int sz[11111111];
int k, n;
int rv[1 << 21];
int a[1 << 21];
int cp = 1;
void pushu(int p) {
  if (s[p][0] && s[p][1])
    dl[p] = dl[s[p][0]], dr[p] = dr[s[p][1]],
    minv[p] =
        min(dr[s[p][0]] + dl[s[p][1]] + 1, min(minv[s[p][0]], minv[s[p][1]]));
  else if (s[p][0])
    dl[p] = dl[s[p][0]], dr[p] = dr[s[p][0]] + sz[p], minv[p] = minv[s[p][0]];
  else if (s[p][1])
    dr[p] = dr[s[p][1]], dl[p] = dl[s[p][1]] + sz[p], minv[p] = minv[s[p][1]];
}
void ins(int x, int p = 1, int i = k - 1) {
  if (i < 0) {
    dl[p] = dr[p] = 0, minv[p] = 1e9;
    return;
  }
  sz[p] = 1 << i;
  int d = (x >> i) & 1;
  if (!s[p][d]) s[p][d] = ++cp;
  ins(x - (d << i), s[p][d], i - 1), pushu(p);
}
void init() {
  for (register int i = 1; i <= n; i++) ins(a[i]);
}
void flip(int dep, int p = 1, int i = k - 1) {
  if (!p) return;
  if (i == dep) {
    swap(s[p][0], s[p][1]), pushu(p);
    return;
  }
  flip(dep, s[p][0], i - 1), flip(dep, s[p][1], i - 1), pushu(p);
}
int T;
int recalc(int S) {
  for (register int i = 0; i < k; i++)
    if (((S ^ T) >> i) & 1) flip(i);
  T = S;
  return minv[1];
}
int ans[1 << 21];
int main() {
  scanf("%d%d", &n, &k);
  for (register int i = 1; i <= n; i++) scanf("%d", a + i);
  init();
  for (register int i = 1; i < (1 << k); i++)
    rv[i] = (rv[i >> 1] >> 1) | ((i & 1) ? (1 << (k - 1)) : 0);
  for (register int i = 0; i < (1 << k); i++) ans[rv[i]] = recalc(rv[i]);
  for (register int i = 0; i < (1 << k); i++) printf("%d ", ans[i]);
  puts("");
}

```