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;
const int NMAX = 1e5;
int v[NMAX + 5], ans[NMAX + 5];
inline void GenerateAnswer(int n) {
  int i, state = 1, st = 1, dr = n, dim = 0;
  sort(v + 1, v + n + 1);
  while (st <= dr) {
    if (st != dr) {
      if (state == 1)
        ans[++dim] = v[st], ans[++dim] = v[dr];
      else
        ans[++dim] = v[dr], ans[++dim] = v[st];
    } else
      ans[++dim] = v[st];
    ++st;
    --dr;
  }
}
int main() {
  int t, n, i, j;
  scanf("%d", &t);
  for (j = 1; j <= t; ++j) {
    scanf("%d", &n);
    for (i = 1; i <= n; ++i) scanf("%d", &v[i]);
    GenerateAnswer(n);
    for (i = n; i >= 1; --i) printf("%d ", ans[i]);
    printf("\n");
  }
  return 0;
}