#include <bits/stdc++.h>
using namespace std;
const int max_k = 19;
const int max_n = (1 << max_k) + 5, inf = 1000111222;
const int max_v = max_n * max_k;
int n, k, a[max_n], ans[max_k][max_n];
int first, nxt[max_v][2];
bool get_bit(int mask, int pos) { return (mask >> pos) & 1; }
void add(int x) {
  int v = 0;
  for (int i = k - 1; i >= 0; --i) {
    if (!nxt[v][get_bit(x, i)]) {
      nxt[v][get_bit(x, i)] = ++first;
    }
    v = nxt[v][get_bit(x, i)];
  }
}
int get_min(int v, int k, int x) {
  int res = 0;
  for (int i = k; i >= 0; --i) {
    int f = get_bit(x, i);
    if (!nxt[v][f]) {
      res ^= 1 << i;
      f ^= 1;
    }
    assert(nxt[v][f]);
    v = nxt[v][f];
  }
  return res;
}
int get_max(int v, int k, int x) {
  int res = 0;
  for (int i = k; i >= 0; --i) {
    int f = !get_bit(x, i);
    res ^= 1 << i;
    if (!nxt[v][f]) {
      res ^= 1 << i;
      f ^= 1;
    }
    assert(nxt[v][f]);
    v = nxt[v][f];
  }
  return res;
}
void solve(int v, int lev) {
  const int max_x = 1 << (lev + 1);
  fill(ans[lev], ans[lev] + max_x, inf);
  if (!lev) {
    if (nxt[v][0] && nxt[v][1]) {
      ans[lev][0] = 1;
      ans[lev][1] = 1;
    }
    return;
  }
  for (int i = 0; i < 2; ++i) {
    if (nxt[v][i]) {
      solve(nxt[v][i], lev - 1);
      for (int x = 0; x < max_x; ++x) {
        ans[lev][x] =
            min(ans[lev][x], ans[lev - 1][x & ((max_x - 1) ^ (1 << lev))]);
      }
    }
  }
  if (nxt[v][0] && nxt[v][1]) {
    for (int x = 0; x < max_x; ++x) {
      const int f = !get_bit(x, lev);
      ans[lev][x] =
          min(ans[lev][x], get_min(nxt[v][f], lev - 1, x) + (1 << lev) -
                               get_max(nxt[v][!f], lev - 1, x));
    }
  }
}
int main() {
  scanf("%d%d", &n, &k);
  for (int i = 0; i < n; ++i) {
    scanf("%d", &a[i]);
    add(a[i]);
  }
  solve(0, k - 1);
  for (int i = 0; i < (1 << k); ++i) {
    printf("%d ", ans[k - 1][i]);
  }
  puts("");
  return 0;
}

### 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 max_k = 19;
const int max_n = (1 << max_k) + 5, inf = 1000111222;
const int max_v = max_n * max_k;
int n, k, a[max_n], ans[max_k][max_n];
int first, nxt[max_v][2];
bool get_bit(int mask, int pos) { return (mask >> pos) & 1; }
void add(int x) {
  int v = 0;
  for (int i = k - 1; i >= 0; --i) {
    if (!nxt[v][get_bit(x, i)]) {
      nxt[v][get_bit(x, i)] = ++first;
    }
    v = nxt[v][get_bit(x, i)];
  }
}
int get_min(int v, int k, int x) {
  int res = 0;
  for (int i = k; i >= 0; --i) {
    int f = get_bit(x, i);
    if (!nxt[v][f]) {
      res ^= 1 << i;
      f ^= 1;
    }
    assert(nxt[v][f]);
    v = nxt[v][f];
  }
  return res;
}
int get_max(int v, int k, int x) {
  int res = 0;
  for (int i = k; i >= 0; --i) {
    int f = !get_bit(x, i);
    res ^= 1 << i;
    if (!nxt[v][f]) {
      res ^= 1 << i;
      f ^= 1;
    }
    assert(nxt[v][f]);
    v = nxt[v][f];
  }
  return res;
}
void solve(int v, int lev) {
  const int max_x = 1 << (lev + 1);
  fill(ans[lev], ans[lev] + max_x, inf);
  if (!lev) {
    if (nxt[v][0] && nxt[v][1]) {
      ans[lev][0] = 1;
      ans[lev][1] = 1;
    }
    return;
  }
  for (int i = 0; i < 2; ++i) {
    if (nxt[v][i]) {
      solve(nxt[v][i], lev - 1);
      for (int x = 0; x < max_x; ++x) {
        ans[lev][x] =
            min(ans[lev][x], ans[lev - 1][x & ((max_x - 1) ^ (1 << lev))]);
      }
    }
  }
  if (nxt[v][0] && nxt[v][1]) {
    for (int x = 0; x < max_x; ++x) {
      const int f = !get_bit(x, lev);
      ans[lev][x] =
          min(ans[lev][x], get_min(nxt[v][f], lev - 1, x) + (1 << lev) -
                               get_max(nxt[v][!f], lev - 1, x));
    }
  }
}
int main() {
  scanf("%d%d", &n, &k);
  for (int i = 0; i < n; ++i) {
    scanf("%d", &a[i]);
    add(a[i]);
  }
  solve(0, k - 1);
  for (int i = 0; i < (1 << k); ++i) {
    printf("%d ", ans[k - 1][i]);
  }
  puts("");
  return 0;
}

```