#include <bits/stdc++.h>
using namespace std;
const int N = 1e6 + 1e3 + 7, INF = 1e9;
int cnt;
struct T {
  int l, r;
  int ls, rs;
  int ld, rd;
  int dis;
  int sz;
} t[N * 2 + 1];
void update(int x) {
  t[x].ld = min({t[t[x].ls].ld, t[t[x].rs].ld + t[t[x].ls].sz, INF});
  t[x].rd = min({t[t[x].rs].rd, t[t[x].ls].rd + t[t[x].rs].sz, INF});
  t[x].sz = t[t[x].ls].sz + t[t[x].rs].sz;
  t[x].dis =
      min({t[t[x].ls].dis, t[t[x].rs].dis, t[t[x].rs].ld + t[t[x].ls].rd});
}
int build(int l, int r) {
  int x = ++cnt;
  t[x].l = l, t[x].r = r;
  if (l == r) {
    t[x].sz = 1;
    t[x].ld = t[x].rd = 1e9;
    t[x].dis = 1e9;
    return x;
  }
  int mid = (l + r) >> 1;
  t[x].ls = build(l, mid);
  t[x].rs = build(mid + 1, r);
  update(x);
  return x;
}
void ins(int x, int v) {
  if (t[x].sz == 1) {
    t[x].ld = t[x].rd = 1;
    return;
  }
  int mid = (t[x].l + t[x].r) >> 1;
  if (v <= mid)
    ins(t[x].ls, v);
  else
    ins(t[x].rs, v);
  update(x);
}
void change(int x, int b, int v) {
  if ((1 << b) & v) {
    swap(t[x].ls, t[x].rs);
    update(x);
    return;
  }
  change(t[x].ls, b - 1, v);
  change(t[x].rs, b - 1, v);
  update(x);
}
vector<int> v;
int ans[N], n, k;
void gen() {
  v.push_back(0);
  for (int i = 1; i <= k; i++) {
    for (auto &x : v) x <<= 1;
    int sz = v.size();
    for (int j = sz - 1; j >= 0; j--) v.push_back(v[j] ^ 1);
  }
}
map<int, int> bts;
int main() {
  scanf("%d%d", &n, &k);
  gen();
  build(0, (1 << k) - 1);
  for (int i = 1; i <= n; i++) {
    int x;
    scanf("%d", &x);
    ins(1, x);
  }
  ans[0] = t[1].dis;
  for (int i = 1; i < v.size(); i++) {
    change(1, k - 1, v[i] ^ v[i - 1]), ans[v[i]] = t[1].dis;
  }
  for (int i = 0; i < (1 << k); i++)
    printf("%d%c", ans[i] - 1, " \n"[i + 1 == 1 << k]);
}

### 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>
using namespace std;
const int N = 1e6 + 1e3 + 7, INF = 1e9;
int cnt;
struct T {
  int l, r;
  int ls, rs;
  int ld, rd;
  int dis;
  int sz;
} t[N * 2 + 1];
void update(int x) {
  t[x].ld = min({t[t[x].ls].ld, t[t[x].rs].ld + t[t[x].ls].sz, INF});
  t[x].rd = min({t[t[x].rs].rd, t[t[x].ls].rd + t[t[x].rs].sz, INF});
  t[x].sz = t[t[x].ls].sz + t[t[x].rs].sz;
  t[x].dis =
      min({t[t[x].ls].dis, t[t[x].rs].dis, t[t[x].rs].ld + t[t[x].ls].rd});
}
int build(int l, int r) {
  int x = ++cnt;
  t[x].l = l, t[x].r = r;
  if (l == r) {
    t[x].sz = 1;
    t[x].ld = t[x].rd = 1e9;
    t[x].dis = 1e9;
    return x;
  }
  int mid = (l + r) >> 1;
  t[x].ls = build(l, mid);
  t[x].rs = build(mid + 1, r);
  update(x);
  return x;
}
void ins(int x, int v) {
  if (t[x].sz == 1) {
    t[x].ld = t[x].rd = 1;
    return;
  }
  int mid = (t[x].l + t[x].r) >> 1;
  if (v <= mid)
    ins(t[x].ls, v);
  else
    ins(t[x].rs, v);
  update(x);
}
void change(int x, int b, int v) {
  if ((1 << b) & v) {
    swap(t[x].ls, t[x].rs);
    update(x);
    return;
  }
  change(t[x].ls, b - 1, v);
  change(t[x].rs, b - 1, v);
  update(x);
}
vector<int> v;
int ans[N], n, k;
void gen() {
  v.push_back(0);
  for (int i = 1; i <= k; i++) {
    for (auto &x : v) x <<= 1;
    int sz = v.size();
    for (int j = sz - 1; j >= 0; j--) v.push_back(v[j] ^ 1);
  }
}
map<int, int> bts;
int main() {
  scanf("%d%d", &n, &k);
  gen();
  build(0, (1 << k) - 1);
  for (int i = 1; i <= n; i++) {
    int x;
    scanf("%d", &x);
    ins(1, x);
  }
  ans[0] = t[1].dis;
  for (int i = 1; i < v.size(); i++) {
    change(1, k - 1, v[i] ^ v[i - 1]), ans[v[i]] = t[1].dis;
  }
  for (int i = 0; i < (1 << k); i++)
    printf("%d%c", ans[i] - 1, " \n"[i + 1 == 1 << k]);
}

```