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 long long maxn = 1e6 + 10;
const long long inf = 0x3f3f3f3f3f3f;
const long long mod = 1e9 + 7;
long long qsm(long long a, long long b, long long mod) {
  long long ans = 1;
  while (b) {
    if (b & 1) ans = ans * a % mod;
    b /= 2;
    a = a * a % mod;
  }
  return ans;
}
long long gcd(long long a, long long b) { return b ? gcd(b, a % b) : a; }
long long T, n, m;
long long a1[maxn], a2[maxn], a3[maxn];
vector<int> q[maxn];
int main() {
  cin >> T;
  while (T--) {
    cin >> n;
    for (int i = 0; i < n; i++) {
      cin >> a1[i];
    }
    sort(a1, a1 + n);
    int left = n / 2 - 1 + n % 2;
    int right = left + 1;
    int now = 0;
    while (now < n) {
      if (now % 2 == 0)
        cout << a1[left--] << " ";
      else
        cout << a1[right++] << " ";
      now++;
    }
    cout << endl;
  }
  return 0;
}