#include <bits/stdc++.h>
using namespace std;
int son[2000010][2] = {0}, k, n, cnt = 1;
int ans[20][2000010];
void ins(int x) {
  int u = 1;
  for (int i = k - 1; i > -1; i--) {
    int v = (x >> i) & 1;
    if (son[u][v] == 0) son[u][v] = ++cnt;
    u = son[u][v];
  }
}
void calc(int id, int d, int x) {
  int t = (x >> (d - 1)) & 1;
  int u = son[id][t], sum1, sum2 = x | (1 << (d - 1));
  sum1 = sum2 ^ (1 << (d - 1));
  for (int i = d - 2; i > -1; i--) {
    int v = (x >> i) & 1;
    if (son[u][v ^ 1]) {
      u = son[u][v ^ 1];
      sum1 |= 1 << i;
    } else {
      u = son[u][v];
      sum1 ^= v << i;
    }
  }
  u = son[id][t ^ 1];
  for (int i = d - 2; i > -1; i--) {
    int v = (x >> i) & 1;
    if (son[u][v]) {
      u = son[u][v];
      sum2 ^= v << i;
    } else {
      u = son[u][v ^ 1];
      sum2 |= 1 << i;
    }
  }
  ans[d][x] = min(ans[d][x], abs(sum1 - sum2));
}
void dfs(int u, int d) {
  if (d <= 0) return;
  if (son[u][0] && son[u][1])
    for (int i = 0; i < 1 << d; i++) calc(u, d, i);
  if (son[u][0]) dfs(son[u][0], d - 1);
  if (son[u][1]) dfs(son[u][1], d - 1);
}
int main() {
  int x;
  memset(ans, 127, sizeof(ans));
  scanf("%d%d", &n, &k);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &x);
    ins(x);
  }
  dfs(1, k);
  for (int i = 0; i < 1 << k; i++) {
    int res = 2e9;
    for (int j = k; j; j--) res = min(res, ans[j][i & ((1 << j) - 1)]);
    printf("%d ", res);
  }
}

### 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 namespace std;
int son[2000010][2] = {0}, k, n, cnt = 1;
int ans[20][2000010];
void ins(int x) {
  int u = 1;
  for (int i = k - 1; i > -1; i--) {
    int v = (x >> i) & 1;
    if (son[u][v] == 0) son[u][v] = ++cnt;
    u = son[u][v];
  }
}
void calc(int id, int d, int x) {
  int t = (x >> (d - 1)) & 1;
  int u = son[id][t], sum1, sum2 = x | (1 << (d - 1));
  sum1 = sum2 ^ (1 << (d - 1));
  for (int i = d - 2; i > -1; i--) {
    int v = (x >> i) & 1;
    if (son[u][v ^ 1]) {
      u = son[u][v ^ 1];
      sum1 |= 1 << i;
    } else {
      u = son[u][v];
      sum1 ^= v << i;
    }
  }
  u = son[id][t ^ 1];
  for (int i = d - 2; i > -1; i--) {
    int v = (x >> i) & 1;
    if (son[u][v]) {
      u = son[u][v];
      sum2 ^= v << i;
    } else {
      u = son[u][v ^ 1];
      sum2 |= 1 << i;
    }
  }
  ans[d][x] = min(ans[d][x], abs(sum1 - sum2));
}
void dfs(int u, int d) {
  if (d <= 0) return;
  if (son[u][0] && son[u][1])
    for (int i = 0; i < 1 << d; i++) calc(u, d, i);
  if (son[u][0]) dfs(son[u][0], d - 1);
  if (son[u][1]) dfs(son[u][1], d - 1);
}
int main() {
  int x;
  memset(ans, 127, sizeof(ans));
  scanf("%d%d", &n, &k);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &x);
    ins(x);
  }
  dfs(1, k);
  for (int i = 0; i < 1 << k; i++) {
    int res = 2e9;
    for (int j = k; j; j--) res = min(res, ans[j][i & ((1 << j) - 1)]);
    printf("%d ", res);
  }
}

```