#include <bits/stdc++.h>
using namespace std;
void fastscan(long long &x) {
  bool neg = false;
  register long long c;
  x = 0;
  c = getchar();
  if (c == '-') {
    neg = true;
    c = getchar();
  }
  for (; (c > 47 && c < 58); c = getchar()) {
    x = x * 10 + c - 48;
  }
  if (neg) {
    x *= -1;
  }
}
void file() {}
long long bs(long long a[], long long s, long long e, long long x) {
  if (s <= e) {
    long long m = (s + e) / 2;
    if (a[m] == x)
      return m;
    else if (a[m] > x)
      return bs(a, s, m - 1, x);
    else
      return bs(a, m + 1, e, x);
  }
  return -1;
}
bool is_pal(string s) { return equal(begin(s), end(s), s.rbegin()); }
long long mypow(long long a, long long b) {
  long long res = 1;
  while (b > 0) {
    if (b & 1) res = (res * a) % 1000000007;
    a = (a * a) % 1000000007;
    b >>= 1;
  }
  return (res % 1000000007);
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  ;
  file();
  long long t;
  cin >> t;
  while (t--) {
    long long n;
    cin >> n;
    ;
    long long a[n];
    for (long long i = 0; i < n; i++) cin >> a[i];
    ;
    sort(a, a + n);
    long long l = 0;
    long long r = n - 1;
    vector<long long> ans;
    while (l <= r) {
      ans.push_back(a[r--]);
      if (r >= l) {
        ans.push_back(a[l++]);
      }
    }
    reverse(ans.begin(), ans.end());
    for (long long i = 0; i < n; i++) cout << ans[i] << " ";
    cout << "\n";
  }
  return 0;
}

### 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;
void fastscan(long long &x) {
  bool neg = false;
  register long long c;
  x = 0;
  c = getchar();
  if (c == '-') {
    neg = true;
    c = getchar();
  }
  for (; (c > 47 && c < 58); c = getchar()) {
    x = x * 10 + c - 48;
  }
  if (neg) {
    x *= -1;
  }
}
void file() {}
long long bs(long long a[], long long s, long long e, long long x) {
  if (s <= e) {
    long long m = (s + e) / 2;
    if (a[m] == x)
      return m;
    else if (a[m] > x)
      return bs(a, s, m - 1, x);
    else
      return bs(a, m + 1, e, x);
  }
  return -1;
}
bool is_pal(string s) { return equal(begin(s), end(s), s.rbegin()); }
long long mypow(long long a, long long b) {
  long long res = 1;
  while (b > 0) {
    if (b & 1) res = (res * a) % 1000000007;
    a = (a * a) % 1000000007;
    b >>= 1;
  }
  return (res % 1000000007);
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  ;
  file();
  long long t;
  cin >> t;
  while (t--) {
    long long n;
    cin >> n;
    ;
    long long a[n];
    for (long long i = 0; i < n; i++) cin >> a[i];
    ;
    sort(a, a + n);
    long long l = 0;
    long long r = n - 1;
    vector<long long> ans;
    while (l <= r) {
      ans.push_back(a[r--]);
      if (r >= l) {
        ans.push_back(a[l++]);
      }
    }
    reverse(ans.begin(), ans.end());
    for (long long i = 0; i < n; i++) cout << ans[i] << " ";
    cout << "\n";
  }
  return 0;
}

```