#include <bits/stdc++.h>
using namespace std;
template <class T>
istream& operator>>(istream& is, vector<T>& v) {
  for (T& x : v) is >> x;
  return is;
}
template <class T>
ostream& operator<<(ostream& os, const vector<T>& v) {
  if (!v.empty()) {
    os << v.front();
    for (int i = 1; i < v.size(); ++i) os << ' ' << v[i];
  }
  return os;
}
const int K = 19;
const int INF = 1e8;
bool vis[2 << K];
int ans[1 << K], myn[1 << K];
void solve(int o, int k, int l) {
  if (k == 0) {
    ans[l] = INF;
    myn[l] = vis[o] ? 0 : INF;
    return;
  }
  solve(o << 1, k - 1, l);
  solve(o << 1 | 1, k - 1, l + (1 << (k - 1)));
  vis[o] = vis[o << 1] | vis[o << 1 | 1];
  for (int i = 0; i != 1 << (k - 1); ++i)
    ans[l + i] = ans[l + i + (1 << (k - 1))] =
        min(ans[l + i], ans[l + i + (1 << (k - 1))]);
  if (vis[o << 1] && vis[o << 1 | 1]) {
    for (int i = 0; i != 1 << (k - 1); ++i) {
      int y = myn[l + i + (1 << (k - 1))],
          x = myn[l + (1 << (k - 1)) - 1 - i] ^ ((1 << (k - 1)) - 1);
      ans[l + i] = min(ans[l + i], y - x + (1 << (k - 1)));
      y = myn[l + i];
      x = myn[l + (1 << k) - 1 - i] ^ ((1 << (k - 1)) - 1);
      ans[l + i + (1 << (k - 1))] =
          min(ans[l + i + (1 << (k - 1))], y - x + (1 << (k - 1)));
    }
  }
  for (int i = 0; i != 1 << (k - 1); ++i) {
    myn[l + i] = min(myn[l + i], myn[l + i + (1 << (k - 1))] + (1 << (k - 1)));
    myn[l + i + (1 << (k - 1))] =
        min(myn[l + i + (1 << (k - 1))], myn[l + i] + (1 << (k - 1)));
  }
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  int n, k;
  cin >> n >> k;
  while (n--) {
    int ai;
    cin >> ai;
    ai += 1 << k;
    vis[ai] = true;
  }
  solve(1, k, 0);
  for (int i = 0; i != 1 << k; ++i) cout << ans[i] << " \n"[i == (1 << k) - 1];
  return 0;
}

### Prompt

In cpp, your task is to solve 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;
template <class T>
istream& operator>>(istream& is, vector<T>& v) {
  for (T& x : v) is >> x;
  return is;
}
template <class T>
ostream& operator<<(ostream& os, const vector<T>& v) {
  if (!v.empty()) {
    os << v.front();
    for (int i = 1; i < v.size(); ++i) os << ' ' << v[i];
  }
  return os;
}
const int K = 19;
const int INF = 1e8;
bool vis[2 << K];
int ans[1 << K], myn[1 << K];
void solve(int o, int k, int l) {
  if (k == 0) {
    ans[l] = INF;
    myn[l] = vis[o] ? 0 : INF;
    return;
  }
  solve(o << 1, k - 1, l);
  solve(o << 1 | 1, k - 1, l + (1 << (k - 1)));
  vis[o] = vis[o << 1] | vis[o << 1 | 1];
  for (int i = 0; i != 1 << (k - 1); ++i)
    ans[l + i] = ans[l + i + (1 << (k - 1))] =
        min(ans[l + i], ans[l + i + (1 << (k - 1))]);
  if (vis[o << 1] && vis[o << 1 | 1]) {
    for (int i = 0; i != 1 << (k - 1); ++i) {
      int y = myn[l + i + (1 << (k - 1))],
          x = myn[l + (1 << (k - 1)) - 1 - i] ^ ((1 << (k - 1)) - 1);
      ans[l + i] = min(ans[l + i], y - x + (1 << (k - 1)));
      y = myn[l + i];
      x = myn[l + (1 << k) - 1 - i] ^ ((1 << (k - 1)) - 1);
      ans[l + i + (1 << (k - 1))] =
          min(ans[l + i + (1 << (k - 1))], y - x + (1 << (k - 1)));
    }
  }
  for (int i = 0; i != 1 << (k - 1); ++i) {
    myn[l + i] = min(myn[l + i], myn[l + i + (1 << (k - 1))] + (1 << (k - 1)));
    myn[l + i + (1 << (k - 1))] =
        min(myn[l + i + (1 << (k - 1))], myn[l + i] + (1 << (k - 1)));
  }
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  int n, k;
  cin >> n >> k;
  while (n--) {
    int ai;
    cin >> ai;
    ai += 1 << k;
    vis[ai] = true;
  }
  solve(1, k, 0);
  for (int i = 0; i != 1 << k; ++i) cout << ans[i] << " \n"[i == (1 << k) - 1];
  return 0;
}

```