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;
int a[100005];
int main() {
  int T;
  scanf("%d", &T);
  for (int t = 1; t <= T; t++) {
    int N;
    scanf("%d", &N);
    int b[100005];
    for (int i = 1; i <= N; i++) {
      scanf("%d", &b[i]);
    }
    sort(b + 1, b + N + 1);
    int mid = N % 2 ? N / 2 + 1 : N / 2;
    int lo = mid, hi = mid;
    if (N % 2 == 0)
      hi++;
    else {
      printf("%d ", b[mid]);
      lo--;
      hi++;
    }
    if (N % 2 == 0) {
      while (lo > 0 && hi <= N) {
        printf("%d %d ", b[lo], b[hi]);
        lo--;
        hi++;
      }
    } else {
      if (b[mid] - b[lo] > b[hi] - b[mid]) {
        while (lo > 0 && hi <= N) {
          printf("%d %d ", b[hi], b[lo]);
          lo--;
          hi++;
        }
      } else {
        while (lo > 0 && hi <= N) {
          printf("%d %d ", b[lo], b[hi]);
          lo--;
          hi++;
        }
      }
    }
    printf("\n");
  }
  return 0;
}