#include <bits/stdc++.h>
using namespace std;
char _c;
bool _f;
template <class T>
inline void IN(T& x) {
  x = 0, _f = 0;
  while (_c = getchar(), !isdigit(_c))
    if (_c == '-') _f = 1;
  while (isdigit(_c)) x = x * 10 + _c - '0', _c = getchar();
  if (_f) x = -x;
}
template <class T>
inline void chkmin(T& x, T y) {
  if (x > y) x = y;
}
template <class T>
inline void chkmax(T& x, T y) {
  if (x < y) x = y;
}
const int N = (1 << 19) | 5;
const int M = N * 20;
const int inf = 1e9 + 7;
int n, k, a[N];
int cnt, mi[M], mx[M], dep[M], siz[M], ans[M], ch[M][2];
inline void pushup(int now) {
  mi[now] = ch[now][0] ? mi[ch[now][0]] : (1 << dep[now]) + mi[ch[now][1]];
  mx[now] = ch[now][1] ? (1 << dep[now]) + mx[ch[now][1]] : mx[ch[now][0]];
  if (siz[now] <= 1)
    ans[now] = inf;
  else {
    if (!ch[now][0] || !ch[now][1])
      ans[now] = ans[ch[now][ch[now][0] == 0]];
    else
      ans[now] = min(min(ans[ch[now][0]], ans[ch[now][1]]),
                     mi[ch[now][1]] + (1 << dep[now]) - mx[ch[now][0]]);
  }
}
void insert(int now, int val) {
  ++siz[now];
  if (dep[now] == -1) return ans[now] = inf, void();
  bool c = (val >> dep[now]) & 1;
  if (!ch[now][c]) ch[now][c] = ++cnt, dep[ch[now][c]] = dep[now] - 1;
  insert(ch[now][c], val), pushup(now);
}
void flip(int now, int val) {
  if (dep[now] == -1 || (!((val >> dep[now]) & 1))) return;
  swap(ch[now][0], ch[now][1]);
  if (ch[now][0]) flip(ch[now][0], val);
  if (ch[now][1]) flip(ch[now][1], val);
  pushup(now);
}
int ANS[N], lst = 0;
void solve(int val, int now) {
  if (now == k) return flip(0, lst ^ val), lst = val, ANS[val] = ans[0], void();
  solve(val, now + 1), solve(val | (1 << now), now + 1);
}
int main() {
  IN(n), IN(k), dep[0] = k - 1;
  for (int i = 1; i <= n; ++i) IN(a[i]), insert(0, a[i]);
  solve(0, 0);
  for (int i = 0; i < 1 << k; ++i) printf("%d ", ANS[i]);
  puts("");
  return 0;
}

### Prompt

Please create a solution in CPP 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>
using namespace std;
char _c;
bool _f;
template <class T>
inline void IN(T& x) {
  x = 0, _f = 0;
  while (_c = getchar(), !isdigit(_c))
    if (_c == '-') _f = 1;
  while (isdigit(_c)) x = x * 10 + _c - '0', _c = getchar();
  if (_f) x = -x;
}
template <class T>
inline void chkmin(T& x, T y) {
  if (x > y) x = y;
}
template <class T>
inline void chkmax(T& x, T y) {
  if (x < y) x = y;
}
const int N = (1 << 19) | 5;
const int M = N * 20;
const int inf = 1e9 + 7;
int n, k, a[N];
int cnt, mi[M], mx[M], dep[M], siz[M], ans[M], ch[M][2];
inline void pushup(int now) {
  mi[now] = ch[now][0] ? mi[ch[now][0]] : (1 << dep[now]) + mi[ch[now][1]];
  mx[now] = ch[now][1] ? (1 << dep[now]) + mx[ch[now][1]] : mx[ch[now][0]];
  if (siz[now] <= 1)
    ans[now] = inf;
  else {
    if (!ch[now][0] || !ch[now][1])
      ans[now] = ans[ch[now][ch[now][0] == 0]];
    else
      ans[now] = min(min(ans[ch[now][0]], ans[ch[now][1]]),
                     mi[ch[now][1]] + (1 << dep[now]) - mx[ch[now][0]]);
  }
}
void insert(int now, int val) {
  ++siz[now];
  if (dep[now] == -1) return ans[now] = inf, void();
  bool c = (val >> dep[now]) & 1;
  if (!ch[now][c]) ch[now][c] = ++cnt, dep[ch[now][c]] = dep[now] - 1;
  insert(ch[now][c], val), pushup(now);
}
void flip(int now, int val) {
  if (dep[now] == -1 || (!((val >> dep[now]) & 1))) return;
  swap(ch[now][0], ch[now][1]);
  if (ch[now][0]) flip(ch[now][0], val);
  if (ch[now][1]) flip(ch[now][1], val);
  pushup(now);
}
int ANS[N], lst = 0;
void solve(int val, int now) {
  if (now == k) return flip(0, lst ^ val), lst = val, ANS[val] = ans[0], void();
  solve(val, now + 1), solve(val | (1 << now), now + 1);
}
int main() {
  IN(n), IN(k), dep[0] = k - 1;
  for (int i = 1; i <= n; ++i) IN(a[i]), insert(0, a[i]);
  solve(0, 0);
  for (int i = 0; i < 1 << k; ++i) printf("%d ", ANS[i]);
  puts("");
  return 0;
}

```