#include <bits/stdc++.h>
using int64 = long long;
const int N = (1 << 19) + 5;
int n, k, a[N];
struct Node {
  int min, max, ans;
};
Node f[2][N];
void work() {
  scanf("%d%d", &n, &k);
  for (int i = 0; i < n; ++i) {
    int x;
    scanf("%d", &x);
    a[x] = 1;
  }
  int cur = 0, nxt = 1;
  for (int i = 0; i < 1 << k; ++i) {
    f[cur][i].ans = N;
    if (a[i]) {
      f[cur][i].min = i;
      f[cur][i].max = i;
    } else {
      f[cur][i].min = N;
      f[cur][i].max = -N;
    }
  }
  for (int j = 0; j < k; ++j) {
    for (int i = 0; i < 1 << k; ++i) {
      Node left = f[cur][i], right = f[cur][i ^ (1 << j)];
      if (i >> j & 1) {
        if (left.min < N) {
          left.min -= 1 << j;
          left.max -= 1 << j;
        }
        if (right.min < N) {
          right.min += 1 << j;
          right.max += 1 << j;
        }
      }
      f[nxt][i].min = std::min(left.min, right.min);
      f[nxt][i].max = std::max(left.max, right.max);
      f[nxt][i].ans = std::min(left.ans, right.ans);
      if (right.min < N && left.max >= 0) {
        f[nxt][i].ans = std::min(f[nxt][i].ans, right.min - left.max);
      }
    }
    std::swap(cur, nxt);
  }
  for (int i = 0; i < 1 << k; ++i) {
    printf("%d%c", f[cur][i].ans, " \n"[i + 1 == (1 << k)]);
  }
}
int main() {
  int T = 1;
  while (T--) {
    work();
  }
  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 int64 = long long;
const int N = (1 << 19) + 5;
int n, k, a[N];
struct Node {
  int min, max, ans;
};
Node f[2][N];
void work() {
  scanf("%d%d", &n, &k);
  for (int i = 0; i < n; ++i) {
    int x;
    scanf("%d", &x);
    a[x] = 1;
  }
  int cur = 0, nxt = 1;
  for (int i = 0; i < 1 << k; ++i) {
    f[cur][i].ans = N;
    if (a[i]) {
      f[cur][i].min = i;
      f[cur][i].max = i;
    } else {
      f[cur][i].min = N;
      f[cur][i].max = -N;
    }
  }
  for (int j = 0; j < k; ++j) {
    for (int i = 0; i < 1 << k; ++i) {
      Node left = f[cur][i], right = f[cur][i ^ (1 << j)];
      if (i >> j & 1) {
        if (left.min < N) {
          left.min -= 1 << j;
          left.max -= 1 << j;
        }
        if (right.min < N) {
          right.min += 1 << j;
          right.max += 1 << j;
        }
      }
      f[nxt][i].min = std::min(left.min, right.min);
      f[nxt][i].max = std::max(left.max, right.max);
      f[nxt][i].ans = std::min(left.ans, right.ans);
      if (right.min < N && left.max >= 0) {
        f[nxt][i].ans = std::min(f[nxt][i].ans, right.min - left.max);
      }
    }
    std::swap(cur, nxt);
  }
  for (int i = 0; i < 1 << k; ++i) {
    printf("%d%c", f[cur][i].ans, " \n"[i + 1 == (1 << k)]);
  }
}
int main() {
  int T = 1;
  while (T--) {
    work();
  }
  return 0;
}

```