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;
long long int answer[300004] = {0};
int main() {
  long long int t;
  cin >> t;
  while (t--) {
    long long int n;
    cin >> n;
    std::vector<long long int> v;
    for (long long int i = 0; i < n; ++i) {
      long long int num;
      cin >> num;
      v.push_back(num);
    }
    sort(v.begin(), v.end());
    long long int pin = 0;
    long long int pend = n - 1;
    long long int k = 0;
    if (pend - pin >= 2) {
      if (abs(v[pend - 1] - v[pin]) > abs(v[pend] - v[pin + 1])) {
        answer[k] = v[pend];
        answer[k + 1] = v[pin];
        answer[k + 2] = v[pend - 1];
        k += 3;
        pend -= 1;
      } else {
        answer[k] = v[pin];
        answer[k + 1] = v[pend];
        answer[k + 2] = v[pin + 1];
        k += 3;
        pin += 1;
      }
    }
    long long int flag = 0;
    long long int pi = 0;
    long long int pe = 0;
    pi = pin + 1;
    pe = pend - 1;
    for (long long int i = k; i < n; ++i) {
      if (pend == n - 2 && flag == 0) {
        answer[i] = v[pi++];
        flag = 1;
      } else if (pend == n - 2 && flag == 1) {
        answer[i] = v[pe--];
        flag = 0;
      } else if (pin == 1 && flag == 0) {
        answer[i] = v[pe--];
        flag = 1;
      } else if (pin == 1 && flag == 1) {
        answer[i] = v[pi++];
        flag = 0;
      }
    }
    reverse(answer, answer + n);
    for (long long int i = 0; i < n; ++i) cout << answer[i] << endl;
  }
  return 0;
}