#include <bits/stdc++.h>
using namespace std;
int j, sz[1050005], rv[1050005], mn[1050005], mx[1050005], v[1050005],
    ans[1050005], c[1050005], ot[1050005];
inline void up(int k) {
  v[k] = v[k << 1] + v[k << 1 | 1];
  if (!v[k]) return;
  if (!v[k << 1 | 1]) {
    mn[k] = mn[k << 1];
    mx[k] = mx[k << 1];
    ans[k] = ans[k << 1];
  } else if (!v[k << 1]) {
    mn[k] = mn[k << 1 | 1] + sz[k << 1];
    mx[k] = mx[k << 1 | 1] + sz[k << 1];
    ans[k] = ans[k << 1 | 1];
  } else {
    mn[k] = mn[k << 1];
    mx[k] = mx[k << 1 | 1] + sz[k << 1];
    ans[k] = min(min(ans[k << 1], ans[k << 1 | 1]),
                 mn[k << 1 | 1] + sz[k << 1] - mx[k << 1]);
  }
}
void build(int k, int l, int r) {
  sz[k] = r - l + 1;
  if (l == r)
    v[k] = c[l], ans[k] = 1e9;
  else
    build(k << 1, l, (l + r >> 1)), build(k << 1 | 1, (l + r >> 1) + 1, r),
        up(k);
}
void update(int k, int l, int r, int x) {
  if (x == j) {
    swap(mn[k << 1], mn[k << 1 | 1]);
    swap(mx[k << 1], mx[k << 1 | 1]);
    swap(ans[k << 1], ans[k << 1 | 1]);
    swap(v[k << 1], v[k << 1 | 1]);
    up(k);
    return;
  }
  ++x;
  update(k << 1, l, (l + r >> 1), x);
  update(k << 1 | 1, (l + r >> 1) + 1, r, x);
  up(k);
}
int n, k, x, i;
int main() {
  cin >> k >> n;
  for (i = 1; i <= k; ++i) cin >> x, c[x] = 1;
  build(1, 0, (1 << n) - 1);
  ot[0] = ans[1];
  for (i = 1; i < (1 << n); ++i) rv[i] = (rv[i >> 1] >> 1) | ((i & 1) << n - 1);
  for (i = 1; i < (1 << n); ot[rv[i]] = ans[1], ++i)
    for (j = 0; j < n; ++j)
      if ((rv[i] ^ rv[i - 1]) & (1 << (n - j - 1)))
        update(1, 0, (1 << n) - 1, 0);
  for (i = 0; i < (1 << n); ++i) cout << ot[i] << "\n";
}

### 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 j, sz[1050005], rv[1050005], mn[1050005], mx[1050005], v[1050005],
    ans[1050005], c[1050005], ot[1050005];
inline void up(int k) {
  v[k] = v[k << 1] + v[k << 1 | 1];
  if (!v[k]) return;
  if (!v[k << 1 | 1]) {
    mn[k] = mn[k << 1];
    mx[k] = mx[k << 1];
    ans[k] = ans[k << 1];
  } else if (!v[k << 1]) {
    mn[k] = mn[k << 1 | 1] + sz[k << 1];
    mx[k] = mx[k << 1 | 1] + sz[k << 1];
    ans[k] = ans[k << 1 | 1];
  } else {
    mn[k] = mn[k << 1];
    mx[k] = mx[k << 1 | 1] + sz[k << 1];
    ans[k] = min(min(ans[k << 1], ans[k << 1 | 1]),
                 mn[k << 1 | 1] + sz[k << 1] - mx[k << 1]);
  }
}
void build(int k, int l, int r) {
  sz[k] = r - l + 1;
  if (l == r)
    v[k] = c[l], ans[k] = 1e9;
  else
    build(k << 1, l, (l + r >> 1)), build(k << 1 | 1, (l + r >> 1) + 1, r),
        up(k);
}
void update(int k, int l, int r, int x) {
  if (x == j) {
    swap(mn[k << 1], mn[k << 1 | 1]);
    swap(mx[k << 1], mx[k << 1 | 1]);
    swap(ans[k << 1], ans[k << 1 | 1]);
    swap(v[k << 1], v[k << 1 | 1]);
    up(k);
    return;
  }
  ++x;
  update(k << 1, l, (l + r >> 1), x);
  update(k << 1 | 1, (l + r >> 1) + 1, r, x);
  up(k);
}
int n, k, x, i;
int main() {
  cin >> k >> n;
  for (i = 1; i <= k; ++i) cin >> x, c[x] = 1;
  build(1, 0, (1 << n) - 1);
  ot[0] = ans[1];
  for (i = 1; i < (1 << n); ++i) rv[i] = (rv[i >> 1] >> 1) | ((i & 1) << n - 1);
  for (i = 1; i < (1 << n); ot[rv[i]] = ans[1], ++i)
    for (j = 0; j < n; ++j)
      if ((rv[i] ^ rv[i - 1]) & (1 << (n - j - 1)))
        update(1, 0, (1 << n) - 1, 0);
  for (i = 0; i < (1 << n); ++i) cout << ot[i] << "\n";
}

```