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 int MOD = (1000 * 1000 * 1000) + 7;
const long long int MOD2 = 998244353;
const long double pi = 3.141592653589793;
long long int power(long long int x, long long int n) {
  long long int result = 1;
  while (n) {
    if (n % 2 == 1) result = result * x;
    n = n / 2;
    x = x * x;
  }
  return result;
}
long long int gcd(long long int a, long long int b) {
  if (!b) return a;
  return gcd(b, a % b);
}
long long int lcm(long long int a, long long int b) {
  return (a * b) / gcd(a, b);
}
long long int bsearch(long long int a[], long long int s, long long int n,
                      long long int val) {
  long long int mid, beg = s, end = n - 1;
  while (beg <= end) {
    mid = (beg + end) / 2;
    if (val == a[mid]) {
      break;
    } else if (val > a[mid]) {
      beg = mid + 1;
    } else {
      end = mid - 1;
    }
  }
  return mid;
}
long long int max_sum_subarray(long long int a[], long long int n) {
  long long int max = 0, s = 0, f = 0, sum = 0, maxs = 0, maxf = 0;
  for (f = 0; f < n; f++) {
    sum += a[f];
    if (sum > max) {
      max = sum;
      maxs = s;
      maxf = f;
    }
    if (sum < 0) {
      sum = 0;
      s = f + 1;
    }
  }
  return max;
}
vector<long long int> prime;
void prime_num(long long int n) {
  vector<long long int> v;
  long long int i, j;
  for (i = 0; i <= n; i++) v.push_back(i);
  v[0] = -1;
  v[1] = -1;
  for (i = 0; i <= sqrt(n); i++) {
    if (v[i] == -1) {
      continue;
    }
    for (j = i + i; j <= n; j += i) {
      v[j] = -1;
    }
  }
  for (i = 2; i <= n; i++) {
    if (v[i] != -1) {
      prime.push_back(i);
    }
  }
}
void io() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
}
int main() {
  io();
  long long int tc;
  cin >> tc;
  while (tc--) {
    long long int n;
    cin >> n;
    long long int i, j;
    long long int a[n];
    for (i = 0; i < n; i++) cin >> a[i];
    sort(a, a + n);
    if (n % 2) cout << a[n / 2] << " ";
    for (i = n / 2 - 1; i >= 0; i--) cout << a[i] << " " << a[n - 1 - i] << " ";
    cout << "\n";
  }
  return 0;
}