#include <bits/stdc++.h>
using namespace std;
void _print(long long t) { cerr << t; }
void _print(int t) { cerr << t; }
void _print(string t) { cerr << t; }
void _print(char t) { cerr << t; }
void _print(long double t) { cerr << t; }
void _print(double t) { cerr << t; }
void _print(unsigned long long t) { cerr << t; }
template <class T, class V>
void _print(pair<T, V> p);
template <class T>
void _print(vector<T> v);
template <class T>
void _print(set<T> v);
template <class T, class V>
void _print(map<T, V> v);
template <class T>
void _print(multiset<T> v);
template <class T, class V>
void _print(pair<T, V> p) {
  cerr << "{";
  _print(p.first);
  cerr << ",";
  _print(p.second);
  cerr << "}";
}
template <class T>
void _print(vector<T> v) {
  cerr << "[ ";
  for (T i : v) {
    _print(i);
    cerr << " ";
  }
  cerr << "]";
}
template <class T>
void _print(set<T> v) {
  cerr << "[ ";
  for (T i : v) {
    _print(i);
    cerr << " ";
  }
  cerr << "]";
}
template <class T>
void _print(multiset<T> v) {
  cerr << "[ ";
  for (T i : v) {
    _print(i);
    cerr << " ";
  }
  cerr << "]";
}
template <class T, class V>
void _print(map<T, V> v) {
  cerr << "[ ";
  for (auto i : v) {
    _print(i);
    cerr << " ";
  }
  cerr << "]";
}
long long cod[1000005] = {0}, sod[1000005] = {0}, pod[1000005] = {0};
long long binpow(long long base, long long power) {
  long long result = 1;
  while (power) {
    if (power % 2) {
      result = result * base;
      power--;
    } else {
      base = base * base;
      power = power / 2;
    }
  }
  return result;
}
void seive() {
  long long i, j;
  bool prime[1000005];
  memset(prime, true, sizeof(prime));
  prime[0] = false;
  prime[1] = false;
  for (i = 2; i * i < 1000005; i++) {
    for (j = i * i; j < 1000005; j = j + i) {
      prime[j] = false;
    }
  }
}
void divisor() {
  long long i, j;
  for (i = 0; i < 1000005; i++) pod[i] = 1;
  for (i = 1; i < 1000005; i++) {
    for (j = i; j < 1000005; j = j + i) {
      cod[j] = (cod[j] + 1) % 1000000007;
      sod[j] = (sod[j] + i) % 1000000007;
      pod[j] = (pod[j] * i) % 1000000007;
    }
  }
}
void solve() {
  long long n, i;
  cin >> n;
  vector<long long> first, last, v(n, 0);
  for (i = 0; i < n; i++) {
    cin >> v[i];
  }
  sort((v).begin(), (v).end());
  for (i = 0; i < n; i++) {
    if (i < (n + 1) / 2)
      first.push_back(v[i]);
    else
      last.push_back(v[i]);
  }
  sort((first).begin(), (first).end(), greater<int>());
  ;
  ;
  for (i = 0; i < n / 2; i++) cout << first[i] << " " << last[i] << " ";
  if (first.size() != last.size()) cout << first[n / 2];
  cout << "\n";
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long t;
  t = 1;
  cin >> t;
  while (t--) {
    solve();
  }
}

### Prompt

Generate a cpp solution to the following 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".

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
void _print(long long t) { cerr << t; }
void _print(int t) { cerr << t; }
void _print(string t) { cerr << t; }
void _print(char t) { cerr << t; }
void _print(long double t) { cerr << t; }
void _print(double t) { cerr << t; }
void _print(unsigned long long t) { cerr << t; }
template <class T, class V>
void _print(pair<T, V> p);
template <class T>
void _print(vector<T> v);
template <class T>
void _print(set<T> v);
template <class T, class V>
void _print(map<T, V> v);
template <class T>
void _print(multiset<T> v);
template <class T, class V>
void _print(pair<T, V> p) {
  cerr << "{";
  _print(p.first);
  cerr << ",";
  _print(p.second);
  cerr << "}";
}
template <class T>
void _print(vector<T> v) {
  cerr << "[ ";
  for (T i : v) {
    _print(i);
    cerr << " ";
  }
  cerr << "]";
}
template <class T>
void _print(set<T> v) {
  cerr << "[ ";
  for (T i : v) {
    _print(i);
    cerr << " ";
  }
  cerr << "]";
}
template <class T>
void _print(multiset<T> v) {
  cerr << "[ ";
  for (T i : v) {
    _print(i);
    cerr << " ";
  }
  cerr << "]";
}
template <class T, class V>
void _print(map<T, V> v) {
  cerr << "[ ";
  for (auto i : v) {
    _print(i);
    cerr << " ";
  }
  cerr << "]";
}
long long cod[1000005] = {0}, sod[1000005] = {0}, pod[1000005] = {0};
long long binpow(long long base, long long power) {
  long long result = 1;
  while (power) {
    if (power % 2) {
      result = result * base;
      power--;
    } else {
      base = base * base;
      power = power / 2;
    }
  }
  return result;
}
void seive() {
  long long i, j;
  bool prime[1000005];
  memset(prime, true, sizeof(prime));
  prime[0] = false;
  prime[1] = false;
  for (i = 2; i * i < 1000005; i++) {
    for (j = i * i; j < 1000005; j = j + i) {
      prime[j] = false;
    }
  }
}
void divisor() {
  long long i, j;
  for (i = 0; i < 1000005; i++) pod[i] = 1;
  for (i = 1; i < 1000005; i++) {
    for (j = i; j < 1000005; j = j + i) {
      cod[j] = (cod[j] + 1) % 1000000007;
      sod[j] = (sod[j] + i) % 1000000007;
      pod[j] = (pod[j] * i) % 1000000007;
    }
  }
}
void solve() {
  long long n, i;
  cin >> n;
  vector<long long> first, last, v(n, 0);
  for (i = 0; i < n; i++) {
    cin >> v[i];
  }
  sort((v).begin(), (v).end());
  for (i = 0; i < n; i++) {
    if (i < (n + 1) / 2)
      first.push_back(v[i]);
    else
      last.push_back(v[i]);
  }
  sort((first).begin(), (first).end(), greater<int>());
  ;
  ;
  for (i = 0; i < n / 2; i++) cout << first[i] << " " << last[i] << " ";
  if (first.size() != last.size()) cout << first[n / 2];
  cout << "\n";
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long t;
  t = 1;
  cin >> t;
  while (t--) {
    solve();
  }
}

```