#include <bits/stdc++.h>
using namespace std;
void rec(vector<int>& mi, int base, int opt, const vector<int>& a, int al,
         int ar, int p) {
  if (p == -1) {
    assert(ar - al == 1);
    int i = opt + 1;
    do {
      i = (i - 1) & opt;
      int x = base + i;
      mi[x] = a[al] ^ x;
    } while (i);
    return;
  }
  int am = ar;
  for (int i = al; i < ar; i++) {
    if (a[i] >> p & 1) {
      am = i;
      break;
    }
  }
  if (am == al) {
    rec(mi, base, opt, a, al, ar, p - 1);
  } else if (am == ar) {
    rec(mi, base, opt, a, al, ar, p - 1);
  } else {
    rec(mi, base, opt ^ (1 << p), a, al, am, p - 1);
    rec(mi, base | 1 << p, opt ^ (1 << p), a, am, ar, p - 1);
  }
}
vector<int> recmi(const vector<int>& a, int p) {
  vector<int> mi(1 << (p + 1));
  rec(mi, 0, (1 << (p + 1)) - 1, a, 0, a.size(), p);
  return mi;
}
vector<int> recma(vector<int> a, int p) {
  for (int& v : a) v ^= (1 << (p + 1)) - 1;
  reverse(a.begin(), a.end());
  vector<int> mi = recmi(a, p);
  for (int& v : mi) v ^= (1 << (p + 1)) - 1;
  return mi;
}
vector<int> slowrecma(vector<int> a, int p) {
  vector<int> ma(1 << (p + 1), -1);
  for (int x = 0; x < 1 << (p + 1); x++)
    for (int v : a) ma[x] = max(ma[x], v ^ x);
  return ma;
}
vector<int> slowrecmi(vector<int> a, int p) {
  vector<int> mi(1 << (p + 1), 1e8);
  for (int x = 0; x < 1 << (p + 1); x++)
    for (int v : a) mi[x] = min(mi[x], v ^ x);
  return mi;
}
int ans[20][1 << 19];
void rec2(const vector<int>& a, int p) {
  if (a.size() <= 1) return;
  assert(p >= 0);
  vector<int> l, r;
  for (int v : a) {
    (v >> p & 1 ? r : l).push_back(v);
  }
  if (l.size() && r.size()) {
    auto lmi = recmi(l, p - 1);
    auto lma = recma(l, p - 1);
    auto rmi = recmi(r, p - 1);
    auto rma = recma(r, p - 1);
    for (int x = 0; x < 1 << (p + 1); x++) {
      if (x >> p & 1) {
        ans[p][x] = min(ans[p][x], ((lmi[x - (1 << p)] + (1 << p)) -
                                    (rma[x - (1 << p)] - (1 << p))));
      } else {
        ans[p][x] = min(ans[p][x], (rmi[x] - lma[x]));
      }
    }
  }
  rec2(l, p - 1);
  rec2(r, p - 1);
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  int n, k;
  cin >> n >> k;
  vector<int> a(n);
  for (int& v : a) cin >> v;
  sort(a.begin(), a.end());
  for (int i = 0; i < k; i++)
    for (int j = 0; j < 1 << k; j++) ans[i][j] = 1e9;
  rec2(a, k - 1);
  for (int x = 0; x < 1 << k; x++) {
    int mi = 1e9;
    for (int i = 0; i < k; i++) mi = min(mi, ans[i][x & ((1 << (i + 1)) - 1)]);
    cout << mi << ' ';
  }
  cout << endl;
}

### 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;
void rec(vector<int>& mi, int base, int opt, const vector<int>& a, int al,
         int ar, int p) {
  if (p == -1) {
    assert(ar - al == 1);
    int i = opt + 1;
    do {
      i = (i - 1) & opt;
      int x = base + i;
      mi[x] = a[al] ^ x;
    } while (i);
    return;
  }
  int am = ar;
  for (int i = al; i < ar; i++) {
    if (a[i] >> p & 1) {
      am = i;
      break;
    }
  }
  if (am == al) {
    rec(mi, base, opt, a, al, ar, p - 1);
  } else if (am == ar) {
    rec(mi, base, opt, a, al, ar, p - 1);
  } else {
    rec(mi, base, opt ^ (1 << p), a, al, am, p - 1);
    rec(mi, base | 1 << p, opt ^ (1 << p), a, am, ar, p - 1);
  }
}
vector<int> recmi(const vector<int>& a, int p) {
  vector<int> mi(1 << (p + 1));
  rec(mi, 0, (1 << (p + 1)) - 1, a, 0, a.size(), p);
  return mi;
}
vector<int> recma(vector<int> a, int p) {
  for (int& v : a) v ^= (1 << (p + 1)) - 1;
  reverse(a.begin(), a.end());
  vector<int> mi = recmi(a, p);
  for (int& v : mi) v ^= (1 << (p + 1)) - 1;
  return mi;
}
vector<int> slowrecma(vector<int> a, int p) {
  vector<int> ma(1 << (p + 1), -1);
  for (int x = 0; x < 1 << (p + 1); x++)
    for (int v : a) ma[x] = max(ma[x], v ^ x);
  return ma;
}
vector<int> slowrecmi(vector<int> a, int p) {
  vector<int> mi(1 << (p + 1), 1e8);
  for (int x = 0; x < 1 << (p + 1); x++)
    for (int v : a) mi[x] = min(mi[x], v ^ x);
  return mi;
}
int ans[20][1 << 19];
void rec2(const vector<int>& a, int p) {
  if (a.size() <= 1) return;
  assert(p >= 0);
  vector<int> l, r;
  for (int v : a) {
    (v >> p & 1 ? r : l).push_back(v);
  }
  if (l.size() && r.size()) {
    auto lmi = recmi(l, p - 1);
    auto lma = recma(l, p - 1);
    auto rmi = recmi(r, p - 1);
    auto rma = recma(r, p - 1);
    for (int x = 0; x < 1 << (p + 1); x++) {
      if (x >> p & 1) {
        ans[p][x] = min(ans[p][x], ((lmi[x - (1 << p)] + (1 << p)) -
                                    (rma[x - (1 << p)] - (1 << p))));
      } else {
        ans[p][x] = min(ans[p][x], (rmi[x] - lma[x]));
      }
    }
  }
  rec2(l, p - 1);
  rec2(r, p - 1);
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  int n, k;
  cin >> n >> k;
  vector<int> a(n);
  for (int& v : a) cin >> v;
  sort(a.begin(), a.end());
  for (int i = 0; i < k; i++)
    for (int j = 0; j < 1 << k; j++) ans[i][j] = 1e9;
  rec2(a, k - 1);
  for (int x = 0; x < 1 << k; x++) {
    int mi = 1e9;
    for (int i = 0; i < k; i++) mi = min(mi, ans[i][x & ((1 << (i + 1)) - 1)]);
    cout << mi << ' ';
  }
  cout << endl;
}

```