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 maxn = 2e5 + 100;
long long a[maxn], b[maxn], c[maxn], x[maxn];
long long f[maxn];
int n;
bool check(long long x) {
  long long mx = a[1], mi = a[1];
  b[1] = a[1];
  for (int i = 2; i <= n; i++) {
    long long p = a[i] + f[x];
    if (a[i] >= b[i - 1]) {
      b[i] = a[i];
      continue;
    } else if (p >= b[i - 1]) {
      b[i] = b[i - 1];
      continue;
    } else
      return 0;
  }
  return 1;
}
int main() {
  int T;
  cin >> T;
  f[1] = 1;
  while (T--) {
    cin >> n;
    for (int i = 1; i <= n; i++) cin >> a[i];
    long long L = 0, R = 40;
    long long ans;
    sort(a + 1, a + 1 + n);
    for (int i = 1, j = n / 2 + 1; i <= n; i += 2, j++) x[i] = a[j];
    for (int i = 2, j = n / 2; i <= n; i += 2, j--) x[i] = a[j];
    for (int i = 1; i <= n; i++) cout << x[i] << " ";
    puts("");
  }
  return 0;
}