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;
double eps = 0.05;
long long mod = 1e9 + 7;
const int INF = 0x3f3f3f3f, inf = 0x3f3f3f3f;
const int MAXN = 2e3 + 10;
const double PI = acos(-1.0);
const int N = 1e6 + 7;
long long a[N];
struct node {
  long long x;
  bool operator<(node &p) { return abs(x - p.x); }
} p[N];
int main() {
  int t;
  scanf("%d", &t);
  while (t--) {
    int n;
    scanf("%d", &n);
    for (int i = 0; i < n; i++) {
      scanf("%lld", &a[i]);
    }
    sort(a, a + n);
    if (n % 2) {
      printf("%lld", a[n / 2]);
      for (int i = 1; i <= n / 2; i++) {
        printf(" %lld %lld", a[n / 2 - i], a[n / 2 + i]);
      }
    } else {
      printf("%lld %lld", a[n / 2 - 1], a[n / 2]);
      for (int i = 1; i < n / 2; i++) {
        printf(" %lld %lld", a[n / 2 - 1 - i], a[n / 2 + i]);
      }
    }
    printf("\n");
  }
  return 0;
}