#include <bits/stdc++.h>
constexpr int inf = 1e9;
struct Node {
  int mn, mx, sz, ans;
  Node *l, *r;
  Node(int val = 0) : ans(inf), sz(1) {
    if (val == 1) {
      mn = mx = 0;
    } else {
      mn = inf;
      mx = -inf;
    }
  }
  void recalc() {
    sz = l->sz + r->sz;
    mn = std::min(l->mn, l->sz + r->mn);
    mx = std::max(l->mx, l->sz + r->mx);
    ans = std::min({l->ans, r->ans, (l->sz + r->mn) - l->mx});
  }
  Node(Node *left, Node *right) : l(left), r(right) { recalc(); }
};
constexpr int K = 19;
int n, k, cnt[1 << K];
Node tree[2 << K];
void build(int p, int l, int r) {
  if (l == r - 1) {
    tree[p] = Node(cnt[l]);
  } else {
    int m = (l + r) / 2;
    build(2 * p, l, m);
    build(2 * p + 1, m, r);
    tree[p] = Node(&tree[2 * p], &tree[2 * p + 1]);
  }
}
void swapSegments(Node *v, int d) {
  if (d == 0) {
    std::swap(v->l, v->r);
    v->recalc();
  } else {
    swapSegments(v->l, d - 1);
    swapSegments(v->r, d - 1);
    v->recalc();
  }
}
int ans[1 << K], cur = 0;
void solve(int d) {
  if (d == k - 1) {
    ans[cur] = tree[1].ans;
  } else {
    solve(d + 1);
  }
  swapSegments(&tree[1], k - 1 - d);
  cur ^= 1 << d;
  if (d == k - 1) {
    ans[cur] = tree[1].ans;
  } else {
    solve(d + 1);
  }
}
int main() {
  std::ios::sync_with_stdio(false);
  std::cin.tie(nullptr);
  std::cin >> n >> k;
  for (int i = 0; i < n; i++) {
    int x;
    std::cin >> x;
    cnt[x]++;
  }
  build(1, 0, 1 << k);
  solve(0);
  for (int i = 0; i < (1 << k); i++) {
    std::cout << ans[i] << " \n"[i == (1 << k) - 1];
  }
  return 0;
}

### Prompt

Your task is to create 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>
constexpr int inf = 1e9;
struct Node {
  int mn, mx, sz, ans;
  Node *l, *r;
  Node(int val = 0) : ans(inf), sz(1) {
    if (val == 1) {
      mn = mx = 0;
    } else {
      mn = inf;
      mx = -inf;
    }
  }
  void recalc() {
    sz = l->sz + r->sz;
    mn = std::min(l->mn, l->sz + r->mn);
    mx = std::max(l->mx, l->sz + r->mx);
    ans = std::min({l->ans, r->ans, (l->sz + r->mn) - l->mx});
  }
  Node(Node *left, Node *right) : l(left), r(right) { recalc(); }
};
constexpr int K = 19;
int n, k, cnt[1 << K];
Node tree[2 << K];
void build(int p, int l, int r) {
  if (l == r - 1) {
    tree[p] = Node(cnt[l]);
  } else {
    int m = (l + r) / 2;
    build(2 * p, l, m);
    build(2 * p + 1, m, r);
    tree[p] = Node(&tree[2 * p], &tree[2 * p + 1]);
  }
}
void swapSegments(Node *v, int d) {
  if (d == 0) {
    std::swap(v->l, v->r);
    v->recalc();
  } else {
    swapSegments(v->l, d - 1);
    swapSegments(v->r, d - 1);
    v->recalc();
  }
}
int ans[1 << K], cur = 0;
void solve(int d) {
  if (d == k - 1) {
    ans[cur] = tree[1].ans;
  } else {
    solve(d + 1);
  }
  swapSegments(&tree[1], k - 1 - d);
  cur ^= 1 << d;
  if (d == k - 1) {
    ans[cur] = tree[1].ans;
  } else {
    solve(d + 1);
  }
}
int main() {
  std::ios::sync_with_stdio(false);
  std::cin.tie(nullptr);
  std::cin >> n >> k;
  for (int i = 0; i < n; i++) {
    int x;
    std::cin >> x;
    cnt[x]++;
  }
  build(1, 0, 1 << k);
  solve(0);
  for (int i = 0; i < (1 << k); i++) {
    std::cout << ans[i] << " \n"[i == (1 << k) - 1];
  }
  return 0;
}

```