#include <bits/stdc++.h>
using namespace std;
const int N = 2e6 + 10, inf = 1e9;
int n, m, tot, rt, ans[N];
struct Node {
  int lc, rc, mn, mx, val;
} t[N];
void up_(int o, int k) {
  t[o].val = t[o].mx = t[o].mn = inf;
  if (t[o].lc) {
    t[o].val = min(t[o].val, t[t[o].lc].val);
    t[o].mn = min(t[o].mn, t[t[o].lc].mn);
    t[o].mx = min(t[o].mx, t[t[o].lc].mx + (1 << k - 1));
  }
  if (t[o].rc) {
    t[o].val = min(t[o].val, t[t[o].rc].val);
    t[o].mn = min(t[o].mn, t[t[o].rc].mn + (1 << k - 1));
    t[o].mx = min(t[o].mx, t[t[o].rc].mx);
  }
  if (t[o].lc && t[o].rc)
    t[o].val = min(t[o].val, t[t[o].lc].mx + t[t[o].rc].mn);
}
void insert_(int &o, int k, int x) {
  if (!o) o = ++tot;
  if (!k) {
    t[o].mn = t[o].mx = 0, t[o].val = inf;
    return;
  }
  if ((x >> k - 1) & 1)
    insert_(t[o].rc, k - 1, x);
  else
    insert_(t[o].lc, k - 1, x);
  up_(o, k);
}
void upd_(int o, int k, int d) {
  if (k == d) {
    swap(t[o].lc, t[o].rc), up_(o, k);
    return;
  }
  if (t[o].lc) upd_(t[o].lc, k - 1, d);
  if (t[o].rc) upd_(t[o].rc, k - 1, d);
  up_(o, k);
}
int main() {
  scanf("%d%d", &n, &m);
  for (int x, i = 1; i <= n; i++) scanf("%d", &x), insert_(rt, m, x);
  ans[0] = t[rt].val;
  for (int i = 1; i < (1 << m); i++) {
    for (int j = 0; j < m; j++)
      if ((i ^ (i - 1)) & (1 << j)) upd_(rt, m, m - j);
    int cur = 0;
    for (int j = 0; j < m; j++)
      if (i & (1 << j)) cur |= 1 << m - j - 1;
    ans[cur] = t[rt].val;
  }
  for (int i = 0; i < (1 << m); i++) printf("%d ", ans[i] + 1);
}

### 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>
using namespace std;
const int N = 2e6 + 10, inf = 1e9;
int n, m, tot, rt, ans[N];
struct Node {
  int lc, rc, mn, mx, val;
} t[N];
void up_(int o, int k) {
  t[o].val = t[o].mx = t[o].mn = inf;
  if (t[o].lc) {
    t[o].val = min(t[o].val, t[t[o].lc].val);
    t[o].mn = min(t[o].mn, t[t[o].lc].mn);
    t[o].mx = min(t[o].mx, t[t[o].lc].mx + (1 << k - 1));
  }
  if (t[o].rc) {
    t[o].val = min(t[o].val, t[t[o].rc].val);
    t[o].mn = min(t[o].mn, t[t[o].rc].mn + (1 << k - 1));
    t[o].mx = min(t[o].mx, t[t[o].rc].mx);
  }
  if (t[o].lc && t[o].rc)
    t[o].val = min(t[o].val, t[t[o].lc].mx + t[t[o].rc].mn);
}
void insert_(int &o, int k, int x) {
  if (!o) o = ++tot;
  if (!k) {
    t[o].mn = t[o].mx = 0, t[o].val = inf;
    return;
  }
  if ((x >> k - 1) & 1)
    insert_(t[o].rc, k - 1, x);
  else
    insert_(t[o].lc, k - 1, x);
  up_(o, k);
}
void upd_(int o, int k, int d) {
  if (k == d) {
    swap(t[o].lc, t[o].rc), up_(o, k);
    return;
  }
  if (t[o].lc) upd_(t[o].lc, k - 1, d);
  if (t[o].rc) upd_(t[o].rc, k - 1, d);
  up_(o, k);
}
int main() {
  scanf("%d%d", &n, &m);
  for (int x, i = 1; i <= n; i++) scanf("%d", &x), insert_(rt, m, x);
  ans[0] = t[rt].val;
  for (int i = 1; i < (1 << m); i++) {
    for (int j = 0; j < m; j++)
      if ((i ^ (i - 1)) & (1 << j)) upd_(rt, m, m - j);
    int cur = 0;
    for (int j = 0; j < m; j++)
      if (i & (1 << j)) cur |= 1 << m - j - 1;
    ans[cur] = t[rt].val;
  }
  for (int i = 0; i < (1 << m); i++) printf("%d ", ans[i] + 1);
}

```