#include <bits/stdc++.h>
using namespace std;
template <class T>
vector<T> readvec(int N) {
  vector<T> res(N);
  for (int i = 0; i < N; ++i) cin >> res[i];
  return res;
}
void one() {
  int N, K;
  cin >> N >> K;
  vector<int> t = readvec<int>(N);
  vector<int> mins(1 << K, 1 << K);
  vector<vector<bool>> exists(K + 1);
  for (int i = 0; i <= K; ++i) {
    exists[i].resize(1 << i, false);
  }
  for (int j : t) exists[K][j] = true;
  for (int k = K - 1; k >= 0; --k) {
    for (int j = 0; j < (1 << k); ++j) {
      exists[k][j] = exists[k + 1][j * 2] | exists[k + 1][j * 2 + 1];
    }
  }
  for (int k = 0; k < K; ++k) {
    for (int i = 0; i < (1 << (K - k - 1)); ++i) {
      int val = 1 << K;
      for (int j = 0; j < (1 << (k + 1)); j += 2) {
        if (exists[k + 1][j] && exists[k + 1][j + 1]) {
          int d = 1;
          int l = j;
          int r = j + 1;
          for (int cb = k + 1; cb < K; ++cb) {
            int currb = ((i >> (K - cb - 1)) & 1);
            d *= 2;
            l *= 2;
            r *= 2;
            if (exists[cb + 1][l ^ 1 ^ currb] && exists[cb + 1][r ^ currb]) {
              r ^= currb;
              l ^= currb ^ 1;
              d -= 1;
            } else if (exists[cb + 1][l] && exists[cb + 1][r]) {
            } else if (exists[cb + 1][l ^ 1] && exists[cb + 1][r ^ 1]) {
              l ^= 1;
              r ^= 1;
            } else {
              r ^= currb ^ 1;
              l ^= currb;
              d += 1;
            }
          }
          assert(d == (r ^ i) - (l ^ i));
          val = min(val, d);
        }
      }
      for (int j = 0; j < (1 << k); ++j) {
        mins[(j << (K - k)) + i] = min(mins[(j << (K - k)) + i], val);
        mins[(j << (K - k)) + (1 << (K - k)) - 1 - i] =
            min(mins[(j << (K - k)) + (1 << (K - k)) - 1 - i], val);
      }
    }
  }
  for (int i = 0; i < (1 << K); ++i) {
    cout << mins[i] << " \n"[i + 1 == (1 << K)];
  }
}
int main() {
  std::ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  one();
  cout << flush;
}

### Prompt

Please provide a Cpp coded solution to the problem described below:

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>
vector<T> readvec(int N) {
  vector<T> res(N);
  for (int i = 0; i < N; ++i) cin >> res[i];
  return res;
}
void one() {
  int N, K;
  cin >> N >> K;
  vector<int> t = readvec<int>(N);
  vector<int> mins(1 << K, 1 << K);
  vector<vector<bool>> exists(K + 1);
  for (int i = 0; i <= K; ++i) {
    exists[i].resize(1 << i, false);
  }
  for (int j : t) exists[K][j] = true;
  for (int k = K - 1; k >= 0; --k) {
    for (int j = 0; j < (1 << k); ++j) {
      exists[k][j] = exists[k + 1][j * 2] | exists[k + 1][j * 2 + 1];
    }
  }
  for (int k = 0; k < K; ++k) {
    for (int i = 0; i < (1 << (K - k - 1)); ++i) {
      int val = 1 << K;
      for (int j = 0; j < (1 << (k + 1)); j += 2) {
        if (exists[k + 1][j] && exists[k + 1][j + 1]) {
          int d = 1;
          int l = j;
          int r = j + 1;
          for (int cb = k + 1; cb < K; ++cb) {
            int currb = ((i >> (K - cb - 1)) & 1);
            d *= 2;
            l *= 2;
            r *= 2;
            if (exists[cb + 1][l ^ 1 ^ currb] && exists[cb + 1][r ^ currb]) {
              r ^= currb;
              l ^= currb ^ 1;
              d -= 1;
            } else if (exists[cb + 1][l] && exists[cb + 1][r]) {
            } else if (exists[cb + 1][l ^ 1] && exists[cb + 1][r ^ 1]) {
              l ^= 1;
              r ^= 1;
            } else {
              r ^= currb ^ 1;
              l ^= currb;
              d += 1;
            }
          }
          assert(d == (r ^ i) - (l ^ i));
          val = min(val, d);
        }
      }
      for (int j = 0; j < (1 << k); ++j) {
        mins[(j << (K - k)) + i] = min(mins[(j << (K - k)) + i], val);
        mins[(j << (K - k)) + (1 << (K - k)) - 1 - i] =
            min(mins[(j << (K - k)) + (1 << (K - k)) - 1 - i], val);
      }
    }
  }
  for (int i = 0; i < (1 << K); ++i) {
    cout << mins[i] << " \n"[i + 1 == (1 << K)];
  }
}
int main() {
  std::ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  one();
  cout << flush;
}

```