#include <bits/stdc++.h>
using namespace std;
const int MAXN = 1 << 19 | 7;
const int Inf = 1e9;
int n, K, Top, A[MAXN];
int F[MAXN], Minv[MAXN], Maxv[MAXN];
int Read() {
  int a = 0, c = 1;
  char b = getchar();
  while (b != '-' && (b < '0' || b > '9')) b = getchar();
  if (b == '-') c = -1, b = getchar();
  while (b >= '0' && b <= '9') a = a * 10 + b - 48, b = getchar();
  return a * c;
}
int Min(int A, int B) { return A < B ? A : B; }
int Max(int A, int B) { return A > B ? A : B; }
int main() {
  n = Read(), K = Read(), Top = (1 << K) - 1;
  for (int i = 0; i <= Top; i++) F[i] = Minv[i] = Inf, Maxv[i] = -Inf;
  for (int i = 1; i <= n; i++) A[i] = Read(), Minv[A[i]] = Maxv[A[i]] = 0;
  for (int Nb = 0, Bit = 1; Nb < K; Nb++, Bit <<= 1)
    for (int Ri = 0, Le; Ri <= Top; Ri++) {
      if (~Ri & Bit) continue;
      Le = Ri ^ Bit, F[Le] = F[Ri] = Min(F[Le], F[Ri]);
      int Minr = Minv[Ri], Maxr = Maxv[Ri], Minl = Minv[Le], Maxl = Maxv[Le];
      F[Le] = Min(F[Le], Minr + Bit - Maxl),
      F[Ri] = Min(F[Ri], Minl + Bit - Maxr);
      Minv[Le] = Min(Minl, Minr + Bit), Maxv[Le] = Max(Maxl, Maxr + Bit);
      Minv[Ri] = Min(Minr, Minl + Bit), Maxv[Ri] = Max(Maxr, Maxl + Bit);
    }
  for (int i = 0; i <= Top; i++) printf("%d ", F[i]);
}

### Prompt

Create a solution in cpp for 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;
const int MAXN = 1 << 19 | 7;
const int Inf = 1e9;
int n, K, Top, A[MAXN];
int F[MAXN], Minv[MAXN], Maxv[MAXN];
int Read() {
  int a = 0, c = 1;
  char b = getchar();
  while (b != '-' && (b < '0' || b > '9')) b = getchar();
  if (b == '-') c = -1, b = getchar();
  while (b >= '0' && b <= '9') a = a * 10 + b - 48, b = getchar();
  return a * c;
}
int Min(int A, int B) { return A < B ? A : B; }
int Max(int A, int B) { return A > B ? A : B; }
int main() {
  n = Read(), K = Read(), Top = (1 << K) - 1;
  for (int i = 0; i <= Top; i++) F[i] = Minv[i] = Inf, Maxv[i] = -Inf;
  for (int i = 1; i <= n; i++) A[i] = Read(), Minv[A[i]] = Maxv[A[i]] = 0;
  for (int Nb = 0, Bit = 1; Nb < K; Nb++, Bit <<= 1)
    for (int Ri = 0, Le; Ri <= Top; Ri++) {
      if (~Ri & Bit) continue;
      Le = Ri ^ Bit, F[Le] = F[Ri] = Min(F[Le], F[Ri]);
      int Minr = Minv[Ri], Maxr = Maxv[Ri], Minl = Minv[Le], Maxl = Maxv[Le];
      F[Le] = Min(F[Le], Minr + Bit - Maxl),
      F[Ri] = Min(F[Ri], Minl + Bit - Maxr);
      Minv[Le] = Min(Minl, Minr + Bit), Maxv[Le] = Max(Maxl, Maxr + Bit);
      Minv[Ri] = Min(Minr, Minl + Bit), Maxv[Ri] = Max(Maxr, Maxl + Bit);
    }
  for (int i = 0; i <= Top; i++) printf("%d ", F[i]);
}

```