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 greater_than(long long arr[], int n, long long x) {
  int l = 0, r = n - 1, ans = -1;
  while (l <= r) {
    int mid = l + (r - l) / 2;
    if (arr[mid] == x) return mid;
    if (arr[mid] < x) l = mid + 1;
    if (arr[mid] > x) {
      ans = mid;
      r = mid - 1;
    }
  }
  return ans;
}
long long floorSqrt(long long x) {
  if (x == 0 || x == 1) return x;
  unsigned long long start = 1, end = x, ans;
  while (start <= end) {
    unsigned long long mid = start + (end - start) / 2;
    if (mid * mid == x) return mid;
    if (mid * mid < x) {
      start = mid + 1;
      ans = mid;
    } else
      end = mid - 1;
  }
  return ans;
}
long long power(int base, int exp) {
  long long res = 1;
  while (exp > 0) {
    if (exp % 2 == 1) res = (res * base) % 1000000007;
    ;
    base = (base * base) % 1000000007;
    ;
    exp /= 2;
  }
  return res % 1000000007;
  ;
}
int main() {
  int q;
  cin >> q;
  while (q--) {
    int n;
    cin >> n;
    long long arr[n];
    for (int i = 0; i < n; i++) {
      cin >> arr[i];
    }
    sort(arr, arr + n);
    vector<pair<int, int>> vec;
    for (int i = 0; i < n / 2; i++) {
      vec.push_back(make_pair(arr[n - 1 - i], arr[i]));
    }
    int x = n / 2;
    if (n % 2 == 0) {
      for (int i = 0; i < x; i++) {
        cout << vec[x - 1 - i].first << " " << vec[x - 1 - i].second << " ";
      }
    } else {
      cout << arr[n / 2] << " ";
      for (int i = 0; i < x; i++) {
        cout << vec[x - 1 - i].first << " " << vec[x - 1 - i].second << " ";
      }
    }
    cout << "\n";
  }
}