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. 
#include <bits/stdc++.h>
using namespace std;
int n, k, gg, c[1 << 20], uu, mus, C[20][1 << 20], a[1 << 20];
set<int> s;
set<int>::iterator it;
void calc(int f1, int f2) {
  int pos = uu;
  mus = 0;
  for (int i = uu; i < k; ++i) {
    if (((f1 >> i) ^ (f2 >> i)) & 1) {
      pos = i + 1;
    }
  }
  for (int i = pos - 2; i >= uu; --i)
    if (((f1 >> i) ^ (f2 >> i)) & 1) {
      mus |= 1 << i;
    }
  for (int x, i = 0; i < (gg << 1); ++i) {
    x = mus | i;
    C[pos][x] = min(C[pos][x], abs((f1 ^ x) - (f2 ^ x)));
  }
}
void add(int l, int r, int x) {
  it = s.lower_bound(l);
  if (it != s.end()) {
    for (int i = c[*it]; a[i] <= r && i <= n; ++i) {
      if (a[i] != x) calc(a[i], x);
    }
  }
}
int i, mx, j, x, r;
int main() {
  cin >> n >> k;
  for (i = 1; i <= n; ++i) cin >> a[i], s.insert(a[i]);
  sort(a + 1, a + n + 1);
  for (i = 1; i <= n; ++i) c[a[i]] = i;
  mx = ((1 << k) / (n - 1));
  gg = 1;
  while (gg < mx) gg <<= 1, ++uu;
  memset(C, 0x3f, sizeof(C));
  for (i = 1; i < (1 << k); ++i) {
    if (c[i]) {
      x = i & ((1 << k) - gg);
      for (j = c[i] - 1; j && a[j] >= x; --j) {
        calc(a[j], i);
      }
      for (j = uu; j < k; ++j) {
        add((i ^ ((1 << j + 1) - gg)) & x,
            (i ^ ((1 << j + 1) - gg) & x) | (gg - 1), i);
      }
    }
  }
  for (i = 1; i < k; ++i) {
    for (j = 0; j < (1 << i); ++j) {
      C[i + 1][j] = min(C[i + 1][j], C[i][j]);
      C[i + 1][1 << i | j] = min(C[i + 1][1 << i | j], C[i][j]);
    }
  }
  for (j = 0; j < (1 << k); ++j) cout << min(mx, C[i][j]) << " ";
}
