#include <bits/stdc++.h>
using namespace std;
using namespace std;
const int dx[] = {-1, -1, -1, 0, 0, 0, 1, 1, 1};
const int dy[] = {-1, 0, 1, -1, 0, 1, -1, 0, 1};
long long gcd(long long x, long long y) { return (!y) ? x : gcd(y, x % y); }
long long lcm(double x, double y) { return ((x / gcd(x, y)) * y); }
long long fact(long long n) { return !n ? 1 : n * fact(n - 1); }
long long nCr(long long n, long long r) {
  long long ans = 1;
  long long div = 1;
  for (long long i = r + 1; i <= n; i++) {
    ans = ans * i;
    ans /= div;
    div++;
  }
  return ans;
}
bool isPrime(long long x) {
  if (x <= 1) return 0;
  for (long long i = 2; i * i <= x; i++) {
    if (x % i == 0) return 0;
  }
  return 1;
}
void A_KHALIFA() {
  std::ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
}
void file() {
  freopen("mex.in", "r", stdin);
  freopen("output.out", "w", stdout);
}
char arr[109][109];
map<char, long long>::iterator it;
int main() {
  A_KHALIFA();
  long long t;
  cin >> t;
  while (t--) {
    long long n;
    cin >> n;
    vector<long long> v(n), ans;
    for (int i = 0; i < n; i++) {
      cin >> v[i];
    }
    sort(v.begin(), v.end());
    int c = 1;
    if (n % 2 == 0) {
      for (int i = 0; i < (n / 2); c++, i++) {
        ans.push_back(v[i]);
        ans.push_back(v[n - c]);
      }
      reverse(ans.begin(), ans.end());
      for (int i = 0; i < ans.size(); i++) {
        cout << ans[i] << " ";
      }
      cout << '\n';
    } else {
      for (int i = 0; i < (n / 2) + 1; c++, i++) {
        ans.push_back(v[i]);
        ans.push_back(v[n - c]);
      }
      reverse(ans.begin(), ans.end());
      for (int i = 1; i < ans.size(); i++) {
        cout << ans[i] << " ";
      }
      cout << '\n';
    }
  }
}

### Prompt

Please provide a cpp coded solution to the problem described below:

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;
using namespace std;
const int dx[] = {-1, -1, -1, 0, 0, 0, 1, 1, 1};
const int dy[] = {-1, 0, 1, -1, 0, 1, -1, 0, 1};
long long gcd(long long x, long long y) { return (!y) ? x : gcd(y, x % y); }
long long lcm(double x, double y) { return ((x / gcd(x, y)) * y); }
long long fact(long long n) { return !n ? 1 : n * fact(n - 1); }
long long nCr(long long n, long long r) {
  long long ans = 1;
  long long div = 1;
  for (long long i = r + 1; i <= n; i++) {
    ans = ans * i;
    ans /= div;
    div++;
  }
  return ans;
}
bool isPrime(long long x) {
  if (x <= 1) return 0;
  for (long long i = 2; i * i <= x; i++) {
    if (x % i == 0) return 0;
  }
  return 1;
}
void A_KHALIFA() {
  std::ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
}
void file() {
  freopen("mex.in", "r", stdin);
  freopen("output.out", "w", stdout);
}
char arr[109][109];
map<char, long long>::iterator it;
int main() {
  A_KHALIFA();
  long long t;
  cin >> t;
  while (t--) {
    long long n;
    cin >> n;
    vector<long long> v(n), ans;
    for (int i = 0; i < n; i++) {
      cin >> v[i];
    }
    sort(v.begin(), v.end());
    int c = 1;
    if (n % 2 == 0) {
      for (int i = 0; i < (n / 2); c++, i++) {
        ans.push_back(v[i]);
        ans.push_back(v[n - c]);
      }
      reverse(ans.begin(), ans.end());
      for (int i = 0; i < ans.size(); i++) {
        cout << ans[i] << " ";
      }
      cout << '\n';
    } else {
      for (int i = 0; i < (n / 2) + 1; c++, i++) {
        ans.push_back(v[i]);
        ans.push_back(v[n - c]);
      }
      reverse(ans.begin(), ans.end());
      for (int i = 1; i < ans.size(); i++) {
        cout << ans[i] << " ";
      }
      cout << '\n';
    }
  }
}

```