#include <bits/stdc++.h>
using namespace std;
template <typename T>
istream &operator>>(istream &is, vector<T> &arr);
template <typename T>
ostream &operator<<(ostream &os, vector<T> &arr);
template <typename T, typename U>
ostream &operator<<(ostream &os, pair<T, U> &p);
template <typename T>
istream &operator>>(istream &is, vector<T> &arr) {
  typename vector<T>::iterator a;
  for (a = arr.begin(); a != arr.end(); a++) {
    is >> *a;
  }
  return is;
}
template <typename T>
ostream &operator<<(ostream &os, vector<T> &arr) {
  typename vector<T>::iterator a;
  for (a = arr.begin(); a != arr.end(); a++) {
    os << *a << " ";
  }
  os << endl;
  return os;
}
template <typename T, typename U>
ostream &operator<<(ostream &os, pair<T, U> &p) {
  os << p.first << " " << p.second << '\n';
  return os;
}
long long INF = 1e18 + 5;
long long MOD = 1e9 + 7;
long long mul(long long a, long long b, long long md = MOD) {
  return (a % md * b % md) % md;
}
long long add(long long a, long long b, long long md = MOD) {
  return (a % md + b % md) % md;
}
bool valid(int i, int j, int n, int m) {
  return i >= 0 && i < n && j >= 0 && j < m;
}
vector<pair<int, int>> dirs = {{1, 0}, {0, 1}, {-1, 0}, {0, -1}};
long long binPow(long long x, long long y) {
  int ans = 1;
  while (y > 0) {
    if (y % 2 == 1) ans = mul(ans, x);
    x = mul(x, x);
    y /= 2;
  }
  return ans;
}
int main() {
  int t;
  cin >> t;
  while (t--) {
    int n;
    cin >> n;
    vector<int> a(n);
    cin >> a;
    sort(a.begin(), a.end());
    stack<int> st;
    int l = 0, r = n - 1;
    bool who = false;
    while (l <= r) {
      if (who) {
        st.push(a[l]);
        l++;
      } else {
        st.push(a[r]);
        r--;
      }
      who ^= 1;
    }
    while (!st.empty()) {
      cout << st.top() << ' ';
      st.pop();
    }
    cout << endl;
  }
}

### Prompt

Please formulate 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;
template <typename T>
istream &operator>>(istream &is, vector<T> &arr);
template <typename T>
ostream &operator<<(ostream &os, vector<T> &arr);
template <typename T, typename U>
ostream &operator<<(ostream &os, pair<T, U> &p);
template <typename T>
istream &operator>>(istream &is, vector<T> &arr) {
  typename vector<T>::iterator a;
  for (a = arr.begin(); a != arr.end(); a++) {
    is >> *a;
  }
  return is;
}
template <typename T>
ostream &operator<<(ostream &os, vector<T> &arr) {
  typename vector<T>::iterator a;
  for (a = arr.begin(); a != arr.end(); a++) {
    os << *a << " ";
  }
  os << endl;
  return os;
}
template <typename T, typename U>
ostream &operator<<(ostream &os, pair<T, U> &p) {
  os << p.first << " " << p.second << '\n';
  return os;
}
long long INF = 1e18 + 5;
long long MOD = 1e9 + 7;
long long mul(long long a, long long b, long long md = MOD) {
  return (a % md * b % md) % md;
}
long long add(long long a, long long b, long long md = MOD) {
  return (a % md + b % md) % md;
}
bool valid(int i, int j, int n, int m) {
  return i >= 0 && i < n && j >= 0 && j < m;
}
vector<pair<int, int>> dirs = {{1, 0}, {0, 1}, {-1, 0}, {0, -1}};
long long binPow(long long x, long long y) {
  int ans = 1;
  while (y > 0) {
    if (y % 2 == 1) ans = mul(ans, x);
    x = mul(x, x);
    y /= 2;
  }
  return ans;
}
int main() {
  int t;
  cin >> t;
  while (t--) {
    int n;
    cin >> n;
    vector<int> a(n);
    cin >> a;
    sort(a.begin(), a.end());
    stack<int> st;
    int l = 0, r = n - 1;
    bool who = false;
    while (l <= r) {
      if (who) {
        st.push(a[l]);
        l++;
      } else {
        st.push(a[r]);
        r--;
      }
      who ^= 1;
    }
    while (!st.empty()) {
      cout << st.top() << ' ';
      st.pop();
    }
    cout << endl;
  }
}

```