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".
#include <bits/stdc++.h>
using namespace std;
void fast() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
}
bool isPrime(long long int n) {
  if (n <= 1) return false;
  if (n <= 3) return true;
  if (n % 2 == 0 || n % 3 == 0) return false;
  for (long long int i = 5; i * i <= n; i = i + 6)
    if (n % i == 0 || n % (i + 2) == 0) return false;
  return true;
}
vector<long long int> primes;
void sieve(long long int n) {
  vector<bool> prime(n + 1, true);
  for (long long int p = 2; p * p <= n; p++) {
    if (prime[p] == true) {
      for (long long int i = p * 2; i <= n; i += p) prime[i] = false;
    }
  }
  for (int p = 2; p <= n; p++)
    if (prime[p]) primes.push_back(p);
}
long long int gcd(long long int x, long long int y) {
  if (y == 0) return x;
  return gcd(y, x % y);
}
long long int lcm(long long int a, long long int b) {
  return (a * b) / gcd(a, b);
}
vector<long long int> vin(long long int n) {
  vector<long long int> a(n);
  for (long long int i = 0; i < n; i++) cin >> a[i];
  return a;
}
void out(vector<long long int> a) {
  for (int i = 0; i < (long long int)a.size(); i++) {
    cout << a[i] << ' ';
  }
  cout << '\n';
}
void in(long long int *a, long long int n) {
  for (long long int i = 0; i < n; i++) cin >> a[i];
}
void out(long long int *a, long long int n) {
  for (long long int i = 0; i < n; i++) {
    cout << a[i] << " ";
  }
}
void out1(vector<long long int> v) {
  for (long long int i = 0; i < (long long int)v.size(); i++) {
    cout << v[i] << " ";
  }
  cout << '\n';
}
int main() {
  long long int t;
  cin >> t;
  while (t--) {
    long long int n;
    cin >> n;
    long long int a[n];
    for (long long int i = 0; i < n; i++) cin >> a[i];
    sort(a, a + n);
    vector<long long int> v1, v2, v3;
    for (long long int i = 0; i < n / 2; i++) {
      v1.push_back(a[i]);
    }
    if (n % 2 == 0) {
      for (long long int i = n / 2; i < n; i++) {
        v2.push_back(a[i]);
      }
    } else {
      for (long long int i = n / 2 + 1; i < n; i++) {
        v2.push_back(a[i]);
      }
    }
    reverse(v2.begin(), v2.end());
    for (long long int i = 0; i < v1.size(); i++) {
      v3.push_back(v2[i]);
      v3.push_back(v1[i]);
    }
    if (n % 2 != 0) {
      v3.push_back(a[n / 2]);
    }
    reverse(v3.begin(), v3.end());
    for (long long int i = 0; i < v3.size(); i++) {
      cout << v3[i] << " ";
    }
    cout << endl;
  }
  return 0;
}
