Problem: You have array of n numbers a_{1}, a_{2}, …, a_{n}. 

Rearrange these numbers to satisfy |a_{1} - a_{2}| ≤ |a_{2} - a_{3}| ≤ … ≤ |a_{n-1} - a_{n}|, where |x| denotes absolute value of x. It's always possible to find such rearrangement.

Note that all numbers in a are not necessarily different. In other words, some numbers of a may be same.

You have to answer independent t test cases.

Input

The first line contains a single integer t (1 ≤ t ≤ 10^{4}) — the number of test cases.

The first line of each test case contains single integer n (3 ≤ n ≤ 10^{5}) — the length of array a. It is guaranteed that the sum of values of n over all test cases in the input does not exceed 10^{5}.

The second line of each test case contains n integers a_{1}, a_{2}, …, a_{n} (-10^{9} ≤ a_{i} ≤ 10^{9}).

Output

For each test case, print the rearranged version of array a which satisfies given condition. If there are multiple valid rearrangements, print any of them.

Example

Input


2
6
5 -2 4 8 6 5
4
8 1 4 2


Output


5 5 4 6 8 -2
1 2 4 8

Note

In the first test case, after given rearrangement, |a_{1} - a_{2}| = 0 ≤ |a_{2} - a_{3}| = 1 ≤ |a_{3} - a_{4}| = 2 ≤ |a_{4} - a_{5}| = 2 ≤ |a_{5} - a_{6}| = 10. There are other possible answers like "5 4 5 6 -2 8".

In the second test case, after given rearrangement, |a_{1} - a_{2}| = 1 ≤ |a_{2} - a_{3}| = 2 ≤ |a_{3} - a_{4}| = 4. There are other possible answers like "2 4 8 1".

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
template <class T>
inline void amin(T &x, const T &y) {
  if (y < x) x = y;
}
template <class T>
inline void amax(T &x, const T &y) {
  if (x < y) x = y;
}
using ll = signed long long;
using PII = pair<int, int>;
using VI = vector<int>;
using VVI = vector<VI>;
const int INF = 0x3f3f3f3f;
const int MOD = 1e9 + 7;
const int MAXN = 700;
const int N = 100005;
int _;
ll n;
int a[N], b[N];
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  for (scanf("%d", &_); _; _--) {
    scanf("%d", &n);
    for (int i = (0); i < (n); i++) scanf("%d", a + i);
    sort(a, a + n);
    for (int i = n - 1, l = 0, r = n - 1; i >= 0; i--) {
      if (i & 1)
        b[i] = a[l++];
      else
        b[i] = a[r--];
    }
    for (int i = (0); i < (n); i++) printf("%d%c", b[i], " \n"[i == n - 1]);
  }
  return 0;
}