#include <bits/stdc++.h>
int max(int a, int b) { return a > b ? a : b; }
int min(int a, int b) { return a < b ? a : b; }
const int L = 1 << 19;
int fi[L], fx[L];
int ans[L];
void powerful_FWT() {
  for (int cur = 1; cur < L; cur <<= 1)
    for (int j = 0; j < L; j += cur << 1)
      for (int k = 0; k < cur; ++k) {
        int x = j + k, y = j + cur + k;
        int tp = min(ans[x], ans[y]);
        ans[x] = min(tp, fi[y] + cur - fx[x]),
        ans[y] = min(tp, fi[x] + cur - fx[y]);
        int nx = min(fi[x], fi[y] + cur), ny = min(fi[y], fi[x] + cur);
        fi[x] = nx, fi[y] = ny;
        nx = max(fx[x], fx[y] + cur), ny = max(fx[y], fx[x] + cur);
        fx[x] = nx, fx[y] = ny;
      }
}
int main() {
  int n, k;
  scanf("%d%d", &n, &k);
  for (int s = 0; s < L; ++s) fi[s] = ans[s] = L, fx[s] = -L;
  for (int i = 1; i <= n; ++i) {
    int x;
    scanf("%d", &x);
    fi[x] = fx[x] = 0;
  }
  powerful_FWT();
  for (int s = 0; s < (1 << k); ++s) printf("%d ", ans[s]);
  return 0;
}

### 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>
int max(int a, int b) { return a > b ? a : b; }
int min(int a, int b) { return a < b ? a : b; }
const int L = 1 << 19;
int fi[L], fx[L];
int ans[L];
void powerful_FWT() {
  for (int cur = 1; cur < L; cur <<= 1)
    for (int j = 0; j < L; j += cur << 1)
      for (int k = 0; k < cur; ++k) {
        int x = j + k, y = j + cur + k;
        int tp = min(ans[x], ans[y]);
        ans[x] = min(tp, fi[y] + cur - fx[x]),
        ans[y] = min(tp, fi[x] + cur - fx[y]);
        int nx = min(fi[x], fi[y] + cur), ny = min(fi[y], fi[x] + cur);
        fi[x] = nx, fi[y] = ny;
        nx = max(fx[x], fx[y] + cur), ny = max(fx[y], fx[x] + cur);
        fx[x] = nx, fx[y] = ny;
      }
}
int main() {
  int n, k;
  scanf("%d%d", &n, &k);
  for (int s = 0; s < L; ++s) fi[s] = ans[s] = L, fx[s] = -L;
  for (int i = 1; i <= n; ++i) {
    int x;
    scanf("%d", &x);
    fi[x] = fx[x] = 0;
  }
  powerful_FWT();
  for (int s = 0; s < (1 << k); ++s) printf("%d ", ans[s]);
  return 0;
}

```