#include <bits/stdc++.h>
using namespace std;
const int K = 19, N = 1 << K, INF = 0x3f3f3f3f;
int n, k, cnt[N], a[N], ans[K][N];
vector<int> b[N], c[N];
void ckmin(int &x, int y) {
  if (y < x) x = y;
}
void calc(vector<int> v, int s, int t) {
  if (v.size() < 2) return;
  if (s < t) {
    for (int i = 0; i < 1 << t; i++)
      ckmin(ans[s][i], abs((v[0] ^ i) - (v[1] ^ i)));
    return;
  }
  for (int i = 0; i < 1 << (s + 1); i++) c[i].clear();
  for (int i = 0; i < 1 << t; i++)
    for (auto u : v) {
      int to = i ^ (1 << s);
      c[to ^ u].push_back(to), to = i ^ (((1 << (s - t)) - 1) << t),
                               c[to ^ u].push_back(to);
    }
  for (int i = 0; i < 1 << (s + 1); i++) {
    int mn = INF;
    sort(c[i].begin(), c[i].end());
    for (int j = 1; j < c[i].size(); j++) ckmin(mn, c[i][j] - c[i][j - 1]);
    ckmin(ans[s][i], mn);
  }
  return;
}
int main() {
  scanf("%d%d", &n, &k);
  for (int i = 0; i < n; i++) scanf("%d", &a[i]);
  int p = 0;
  for (int i = 1; i <= k; i++) {
    int mx = 0;
    for (int j = 0; j < 1 << k; j++) cnt[j] = 0;
    for (int j = 0; j < n; j++) cnt[a[j] >> i]++;
    for (int j = 0; j < 1 << k; j++) mx = max(mx, cnt[j]);
    if (mx > 1) {
      p = i;
      break;
    }
  }
  memset(ans, 0x3f, sizeof(ans));
  for (int i = p - 1; i < k; i++) {
    for (int j = 0; j < 1 << (k - i - 1); j++) b[j].clear();
    for (int j = 0; j < n; j++)
      b[a[j] >> (i + 1)].push_back(a[j] & ((1 << (i + 1)) - 1));
    for (int j = 0; j < 1 << (k - i - 1); j++) calc(b[j], i, p);
  }
  for (int i = 0; i < 1 << k; i++) {
    int mn = INF;
    for (int j = p - 1; j < k; j++) ckmin(mn, ans[j][i & ((1 << (j + 1)) - 1)]);
    printf("%d ", mn);
  }
}

### Prompt

Your challenge is to write 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>
using namespace std;
const int K = 19, N = 1 << K, INF = 0x3f3f3f3f;
int n, k, cnt[N], a[N], ans[K][N];
vector<int> b[N], c[N];
void ckmin(int &x, int y) {
  if (y < x) x = y;
}
void calc(vector<int> v, int s, int t) {
  if (v.size() < 2) return;
  if (s < t) {
    for (int i = 0; i < 1 << t; i++)
      ckmin(ans[s][i], abs((v[0] ^ i) - (v[1] ^ i)));
    return;
  }
  for (int i = 0; i < 1 << (s + 1); i++) c[i].clear();
  for (int i = 0; i < 1 << t; i++)
    for (auto u : v) {
      int to = i ^ (1 << s);
      c[to ^ u].push_back(to), to = i ^ (((1 << (s - t)) - 1) << t),
                               c[to ^ u].push_back(to);
    }
  for (int i = 0; i < 1 << (s + 1); i++) {
    int mn = INF;
    sort(c[i].begin(), c[i].end());
    for (int j = 1; j < c[i].size(); j++) ckmin(mn, c[i][j] - c[i][j - 1]);
    ckmin(ans[s][i], mn);
  }
  return;
}
int main() {
  scanf("%d%d", &n, &k);
  for (int i = 0; i < n; i++) scanf("%d", &a[i]);
  int p = 0;
  for (int i = 1; i <= k; i++) {
    int mx = 0;
    for (int j = 0; j < 1 << k; j++) cnt[j] = 0;
    for (int j = 0; j < n; j++) cnt[a[j] >> i]++;
    for (int j = 0; j < 1 << k; j++) mx = max(mx, cnt[j]);
    if (mx > 1) {
      p = i;
      break;
    }
  }
  memset(ans, 0x3f, sizeof(ans));
  for (int i = p - 1; i < k; i++) {
    for (int j = 0; j < 1 << (k - i - 1); j++) b[j].clear();
    for (int j = 0; j < n; j++)
      b[a[j] >> (i + 1)].push_back(a[j] & ((1 << (i + 1)) - 1));
    for (int j = 0; j < 1 << (k - i - 1); j++) calc(b[j], i, p);
  }
  for (int i = 0; i < 1 << k; i++) {
    int mn = INF;
    for (int j = p - 1; j < k; j++) ckmin(mn, ans[j][i & ((1 << (j + 1)) - 1)]);
    printf("%d ", mn);
  }
}

```