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;
int siftp(int *a, int *p, int maxn) {
  int pn = 0;
  a[0] = a[1] = 1;
  for (int i = 2; i <= maxn; i++) {
    if (!a[i]) p[pn++] = i;
    for (int j = 0; j < pn; j++) {
      if (i * p[j] > maxn) break;
      a[i * p[j]] = 1;
      if (i % p[j] == 0) break;
    }
  }
  return pn;
}
bool isprime(int x) {
  if (x < 2) return false;
  for (int i = 2; i * i <= x; i++)
    if (x % i == 0) return false;
  return true;
}
long long gcd(long long a, long long b) { return b ? gcd(b, a % b) : a; }
long long powmod(long long a, long long b, long long c) {
  if (0 == b) return 1;
  long long x = powmod(a, b / 2, c);
  x = x * x % c;
  if (b & 1) x = x * a % c;
  return x;
}
int n, a[1005];
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n; i++) scanf("%d", a + i);
  sort(a, a + n);
  for (int i = 0, j = n - 1, k = 0; i <= j; k++) {
    if (k & 1) {
      printf("%d ", a[j--]);
    } else {
      printf("%d ", a[i++]);
    }
  }
  return 0;
}
