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".

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const long long mod = 1e9 + 7;
const long long INF = 0x3f3f3f3f3f3f3f3f;
const int inf = 0x3f3f3f3f;
const int maxn = 2e5 + 5;
const double eps = 1e-8;
const double pi = acos(-1.0);
long long gcd(long long a, long long b) { return !b ? a : gcd(b, a % b); }
long long lcm(long long a, long long b) { return a * b / gcd(a, b); }
long long poww(long long x, long long y, long long p) {
  long long ans = 1;
  while (y) {
    if (y & 1) ans = ans * x % p;
    x = x * x % p;
    y >>= 1;
  }
  return ans % p;
}
bool isp(long long x) {
  if (x < 2) return 0;
  for (long long i = 2; i * i <= x; i++) {
    if (x % i == 0) return 0;
  }
  return 1;
}
long long t, n, a[maxn], ans[maxn];
void solve() {
  if (n & 1) {
    ans[0] = a[n / 2];
    int l = n / 2, r = n / 2, cur = 0;
    while (l > 0 || r < n) {
      int x = a[--l], y = a[++r];
      if (abs(ans[cur] - x) > ans[cur] - y) swap(x, y);
      ans[++cur] = x;
      ans[++cur] = y;
    }
  } else {
    int l = n / 2 - 1, r = n / 2, cur = 0;
    while (l > 0 || r < n) {
      int x = a[l--], y = a[r++];
      if (abs(ans[cur] - x) > ans[cur] - y) swap(x, y);
      ans[cur++] = x;
      ans[cur++] = y;
    }
  }
  for (int i = (0); i <= (n - 1); i++) cout << ans[i] << " ";
  cout << endl;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> t;
  while (t--) {
    cin >> n;
    for (int i = 0; i < n; i++) cin >> a[i];
    sort(a, a + n);
    solve();
  }
  return 0;
}