#include <bits/stdc++.h>
const int K = 19;
const int N = 1 << K;
const int S = N << 2;
int n, k, a[N];
bool tag[N];
int order[N], ans[N];
int lv[S], rv[S], dis[S], len[S];
int ch[S][2], total = 0, root;
void pushup(int x) {
  len[x] = len[ch[x][0]] + len[ch[x][1]];
  lv[x] = std::min(lv[ch[x][0]], lv[ch[x][1]] + len[ch[x][0]]);
  rv[x] = std::max(rv[ch[x][0]], rv[ch[x][1]] + len[ch[x][0]]);
  dis[x] = std::min({dis[ch[x][0]], dis[ch[x][1]],
                     lv[ch[x][1]] + len[ch[x][0]] - rv[ch[x][0]]});
}
int build(int l, int r) {
  int x = ++total;
  if (l + 1 == r) {
    len[x] = 1;
    if (tag[l]) lv[x] = rv[x] = 0;
    return x;
  }
  ch[x][0] = build(l, (l + r) / 2);
  ch[x][1] = build((l + r) / 2, r);
  return pushup(x), x;
}
void flip(int x, int tar, int height = k - 1) {
  if (height == tar) {
    std::swap(ch[x][0], ch[x][1]);
  } else {
    flip(ch[x][0], tar, height - 1);
    flip(ch[x][1], tar, height - 1);
  }
  pushup(x);
}
signed main() {
  memset(lv, 0x3f, sizeof(lv));
  memset(rv, -0x3f, sizeof(rv));
  memset(dis, 0x3f, sizeof(dis));
  scanf("%d%d", &n, &k);
  for (int i = 0; i < n; i++) scanf("%d", a + i), tag[a[i]] = 1;
  for (int i = 0; i < (1 << k); i++) order[i] = i ^ (i >> 1);
  for (int i = 0; i < (1 << k); i++) {
    int cur = 0;
    for (int j = 0; j < k; j++) cur = cur << 1 | (order[i] >> j & 1);
    order[i] = cur;
  }
  root = build(0, 1 << k);
  ans[0] = dis[root];
  for (int i = 1; i < (1 << k); i++) {
    int layer_id = __builtin_ctz(order[i] ^ order[i - 1]);
    flip(root, layer_id), ans[order[i]] = dis[root];
  }
  for (int i = 0; i < (1 << k); i++)
    printf("%d%c", ans[i], " \n"[i == (1 << k) - 1]);
  return 0;
}

### Prompt

Generate 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>
const int K = 19;
const int N = 1 << K;
const int S = N << 2;
int n, k, a[N];
bool tag[N];
int order[N], ans[N];
int lv[S], rv[S], dis[S], len[S];
int ch[S][2], total = 0, root;
void pushup(int x) {
  len[x] = len[ch[x][0]] + len[ch[x][1]];
  lv[x] = std::min(lv[ch[x][0]], lv[ch[x][1]] + len[ch[x][0]]);
  rv[x] = std::max(rv[ch[x][0]], rv[ch[x][1]] + len[ch[x][0]]);
  dis[x] = std::min({dis[ch[x][0]], dis[ch[x][1]],
                     lv[ch[x][1]] + len[ch[x][0]] - rv[ch[x][0]]});
}
int build(int l, int r) {
  int x = ++total;
  if (l + 1 == r) {
    len[x] = 1;
    if (tag[l]) lv[x] = rv[x] = 0;
    return x;
  }
  ch[x][0] = build(l, (l + r) / 2);
  ch[x][1] = build((l + r) / 2, r);
  return pushup(x), x;
}
void flip(int x, int tar, int height = k - 1) {
  if (height == tar) {
    std::swap(ch[x][0], ch[x][1]);
  } else {
    flip(ch[x][0], tar, height - 1);
    flip(ch[x][1], tar, height - 1);
  }
  pushup(x);
}
signed main() {
  memset(lv, 0x3f, sizeof(lv));
  memset(rv, -0x3f, sizeof(rv));
  memset(dis, 0x3f, sizeof(dis));
  scanf("%d%d", &n, &k);
  for (int i = 0; i < n; i++) scanf("%d", a + i), tag[a[i]] = 1;
  for (int i = 0; i < (1 << k); i++) order[i] = i ^ (i >> 1);
  for (int i = 0; i < (1 << k); i++) {
    int cur = 0;
    for (int j = 0; j < k; j++) cur = cur << 1 | (order[i] >> j & 1);
    order[i] = cur;
  }
  root = build(0, 1 << k);
  ans[0] = dis[root];
  for (int i = 1; i < (1 << k); i++) {
    int layer_id = __builtin_ctz(order[i] ^ order[i - 1]);
    flip(root, layer_id), ans[order[i]] = dis[root];
  }
  for (int i = 0; i < (1 << k); i++)
    printf("%d%c", ans[i], " \n"[i == (1 << k) - 1]);
  return 0;
}

```