#include <bits/stdc++.h>
template <typename _Tp>
void read(_Tp &x) {
  char ch(getchar());
  bool f(false);
  while (!isdigit(ch)) f |= ch == 45, ch = getchar();
  x = ch & 15, ch = getchar();
  while (isdigit(ch)) x = x * 10 + (ch & 15), ch = getchar();
  if (f) x = -x;
}
template <typename _Tp, typename... Args>
void read(_Tp &t, Args &...args) {
  read(t);
  read(args...);
}
const int N = (1 << 19) + 5, inf = 1e9;
int pool[N * 20 * 3], *P = pool;
int *mn[N << 1], *mx[N << 1], *ans[N << 1], dep[N << 1];
bool _[N];
void build(int l, int r, int u) {
  mn[u] = P, P += r - l + 1, mx[u] = P, P += r - l + 1, ans[u] = P,
  P += r - l + 1;
  if (l == r) {
    if (_[l])
      mn[u][0] = 0, mx[u][0] = 0, ans[u][0] = inf;
    else
      mn[u][0] = inf, mx[u][0] = -inf, ans[u][0] = inf;
    return;
  }
  int mid = (l + r) >> 1;
  dep[((u) << 1)] = dep[u] - 1, dep[((u) << 1 | 1)] = dep[u] - 1;
  build(l, mid, ((u) << 1)), build(mid + 1, r, ((u) << 1 | 1));
  int ch[2] = {((u) << 1), ((u) << 1 | 1)}, M = (1 << (dep[u] - 1)) - 1;
  for (int i = 0; i < 1 << dep[u]; ++i) {
    int o = i >> (dep[u] - 1) & 1;
    mn[u][i] =
        std::min(mn[ch[o]][i & M], mn[ch[!o]][i & M] + (1 << (dep[u] - 1)));
    mx[u][i] =
        std::max(mx[ch[o]][i & M], mx[ch[!o]][i & M] + (1 << (dep[u] - 1)));
    ans[u][i] = std::min(
        {ans[ch[0]][i & M], ans[ch[1]][i & M],
         abs(mx[ch[o]][i & M] - (mn[ch[!o]][i & M] + (1 << (dep[u] - 1))))});
  }
}
int main() {
  int n, k;
  read(n, k);
  for (int i = 1, x; i <= n; ++i) read(x), _[x] = true;
  dep[1] = k, build(0, (1 << k) - 1, 1);
  for (int i = 0; i < 1 << k; ++i)
    printf("%d%c", ans[1][i], " \n"[i + 1 == (1 << k)]);
  return 0;
}

### Prompt

Develop a solution in cpp 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>
template <typename _Tp>
void read(_Tp &x) {
  char ch(getchar());
  bool f(false);
  while (!isdigit(ch)) f |= ch == 45, ch = getchar();
  x = ch & 15, ch = getchar();
  while (isdigit(ch)) x = x * 10 + (ch & 15), ch = getchar();
  if (f) x = -x;
}
template <typename _Tp, typename... Args>
void read(_Tp &t, Args &...args) {
  read(t);
  read(args...);
}
const int N = (1 << 19) + 5, inf = 1e9;
int pool[N * 20 * 3], *P = pool;
int *mn[N << 1], *mx[N << 1], *ans[N << 1], dep[N << 1];
bool _[N];
void build(int l, int r, int u) {
  mn[u] = P, P += r - l + 1, mx[u] = P, P += r - l + 1, ans[u] = P,
  P += r - l + 1;
  if (l == r) {
    if (_[l])
      mn[u][0] = 0, mx[u][0] = 0, ans[u][0] = inf;
    else
      mn[u][0] = inf, mx[u][0] = -inf, ans[u][0] = inf;
    return;
  }
  int mid = (l + r) >> 1;
  dep[((u) << 1)] = dep[u] - 1, dep[((u) << 1 | 1)] = dep[u] - 1;
  build(l, mid, ((u) << 1)), build(mid + 1, r, ((u) << 1 | 1));
  int ch[2] = {((u) << 1), ((u) << 1 | 1)}, M = (1 << (dep[u] - 1)) - 1;
  for (int i = 0; i < 1 << dep[u]; ++i) {
    int o = i >> (dep[u] - 1) & 1;
    mn[u][i] =
        std::min(mn[ch[o]][i & M], mn[ch[!o]][i & M] + (1 << (dep[u] - 1)));
    mx[u][i] =
        std::max(mx[ch[o]][i & M], mx[ch[!o]][i & M] + (1 << (dep[u] - 1)));
    ans[u][i] = std::min(
        {ans[ch[0]][i & M], ans[ch[1]][i & M],
         abs(mx[ch[o]][i & M] - (mn[ch[!o]][i & M] + (1 << (dep[u] - 1))))});
  }
}
int main() {
  int n, k;
  read(n, k);
  for (int i = 1, x; i <= n; ++i) read(x), _[x] = true;
  dep[1] = k, build(0, (1 << k) - 1, 1);
  for (int i = 0; i < 1 << k; ++i)
    printf("%d%c", ans[1][i], " \n"[i + 1 == (1 << k)]);
  return 0;
}

```