A student of z-school found a kind of sorting called z-sort. The array a with n elements are z-sorted if two conditions hold:

  1. ai ≥ ai - 1 for all even i, 
  2. ai ≤ ai - 1 for all odd i > 1. 



For example the arrays [1,2,1,2] and [1,1,1,1] are z-sorted while the array [1,2,3,4] isn’t z-sorted.

Can you make the array z-sorted?

Input

The first line contains a single integer n (1 ≤ n ≤ 1000) — the number of elements in the array a.

The second line contains n integers ai (1 ≤ ai ≤ 109) — the elements of the array a.

Output

If it's possible to make the array a z-sorted print n space separated integers ai — the elements after z-sort. Otherwise print the only word "Impossible".

Examples

Input

4
1 2 2 1


Output

1 2 1 2


Input

5
1 3 2 2 5


Output

1 5 2 3 2
#include <bits/stdc++.h>
using namespace std;
long long power(long long a, long long b) {
  if (b == 0) return (long long)1;
  long long x = power(a, b / 2);
  x = x % 1000000007;
  long long ans =
      ((long long)(x % 1000000007) * (long long)(x % 1000000007)) % 1000000007;
  if (b % 2 != 0)
    ans = ((long long)(ans % 1000000007) * (long long)(a % 1000000007)) %
          1000000007;
  ans = ans % 1000000007;
  return ans;
}
long long mod_inv(long long a) {
  long long ans = power(a, 1000000007 - 2);
  ans = ans % 1000000007;
  return ans;
}
long long gcd(long long a, long long b) {
  if (a == 0) return b;
  if (b == 0) return a;
  if (a == b) return a;
  if (a > b)
    return gcd(b, a % b);
  else
    return gcd(a, b % a);
}
void func() {
  long long n;
  cin >> n;
  vector<long long> arr(n, 0);
  for (long long i = 0; i < n; i++) cin >> arr[i];
  sort(arr.begin(), arr.end());
  long long i = 0;
  long long j = n - 1;
  long long cnt = 0;
  while (i <= j) {
    if (cnt % 2 == 0)
      cout << arr[i++] << " ";
    else
      cout << arr[j--] << " ";
    cnt++;
  }
}
int main() {
  long long T = 1;
  while (T--) {
    func();
  }
}
