#include <bits/stdc++.h>
using namespace std;
template <typename T>
T lcm(T a, T b) {
  return a / __gcd(a, b) * b;
}
template <class T>
inline T bigmod(T p, T e, T M) {
  T ret = 1;
  for (; e > 0; e >>= 1) {
    if (e & 1) ret = (ret * p) % M;
    p = (p * p) % M;
  }
  return (T)ret;
}
template <class string>
long long int strton(string s) {
  long long int won;
  stringstream ss;
  ss << s;
  ss >> won;
  return won;
}
template <class T>
string ntostr(T n) {
  stringstream ss;
  ss << n;
  return ss.str();
}
template <class T>
inline T mod(T n, T m) {
  return (n % m + m) % m;
}
template <class T>
bool isprime(T n) {
  if (n % 2 == 0) {
    if (n == 2)
      return true;
    else
      return false;
  }
  T m = sqrt(n);
  for (T i = 3; i <= m; i += 2) {
    if (n % i == 0) return false;
  }
  return true;
}
bool isvowel(char ch) {
  ch = toupper(ch);
  if (ch == 'A' || ch == 'U' || ch == 'I' || ch == 'O' || ch == 'E')
    return true;
  return false;
}
double triangle_area1(double a, double b, double c) {
  double s = (a + b + c) * 0.5;
  return sqrt(s * (s - a) * (s - b) * (s - c));
}
double triangle_area2(double x1, double y1, double x2, double y2, double x3,
                      double y3) {
  return 0.5 * abs(x1 * y2 + x2 * y3 + x3 * y1 - x2 * y1 - x3 * y2 - x1 * y3);
}
double circle_area(double r, double theta) {
  return (theta * (2.0 * acos(0.0)) * r * r) / (2 * (2.0 * acos(0.0)));
}
double quad_area(double x1, double y1, double x2, double y2, double x3,
                 double y3, double x4, double y4) {
  return 0.5 * abs(x1 * y2 + x2 * y3 + x3 * y4 + x4 * y1 - x2 * y1 - x3 * y2 -
                   x4 * y3 - x1 * y4);
}
double cosangle(double a, double b, double c) {
  return acos((double)(b * b + c * c - a * a) / (double)(2 * b * c));
}
long long int Set(long long int N, long long int pos) {
  return N = N | (1ll << pos);
}
long long int reset(long long int N, long long int pos) {
  return N = N & ~(1ll << pos);
}
bool check(long long int N, long long int pos) {
  return (bool)(N & (1ll << pos));
}
long long int a[100005];
vector<long long int> ans;
int main() {
  int t, cno = 0;
  scanf("%d", &t);
  ;
  while (t--) {
    ans.clear();
    long long int n;
    cin >> n;
    for (long long int i = 0; i < n; i++) {
      cin >> a[i];
    }
    sort(a, a + n);
    int i = n / 2;
    i--;
    int j = i + 1;
    if (n % 2) {
      ans.push_back(a[j]);
      j++;
    }
    for (; i >= 0; i--, j++) {
      ans.push_back(a[i]);
      ans.push_back(a[j]);
    }
    for (long long int i = 0; i < n; i++) {
      if (i) {
        cout << " ";
      }
      cout << ans[i];
    }
    cout << '\n';
  }
}

### Prompt

Your challenge is to write 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>
T lcm(T a, T b) {
  return a / __gcd(a, b) * b;
}
template <class T>
inline T bigmod(T p, T e, T M) {
  T ret = 1;
  for (; e > 0; e >>= 1) {
    if (e & 1) ret = (ret * p) % M;
    p = (p * p) % M;
  }
  return (T)ret;
}
template <class string>
long long int strton(string s) {
  long long int won;
  stringstream ss;
  ss << s;
  ss >> won;
  return won;
}
template <class T>
string ntostr(T n) {
  stringstream ss;
  ss << n;
  return ss.str();
}
template <class T>
inline T mod(T n, T m) {
  return (n % m + m) % m;
}
template <class T>
bool isprime(T n) {
  if (n % 2 == 0) {
    if (n == 2)
      return true;
    else
      return false;
  }
  T m = sqrt(n);
  for (T i = 3; i <= m; i += 2) {
    if (n % i == 0) return false;
  }
  return true;
}
bool isvowel(char ch) {
  ch = toupper(ch);
  if (ch == 'A' || ch == 'U' || ch == 'I' || ch == 'O' || ch == 'E')
    return true;
  return false;
}
double triangle_area1(double a, double b, double c) {
  double s = (a + b + c) * 0.5;
  return sqrt(s * (s - a) * (s - b) * (s - c));
}
double triangle_area2(double x1, double y1, double x2, double y2, double x3,
                      double y3) {
  return 0.5 * abs(x1 * y2 + x2 * y3 + x3 * y1 - x2 * y1 - x3 * y2 - x1 * y3);
}
double circle_area(double r, double theta) {
  return (theta * (2.0 * acos(0.0)) * r * r) / (2 * (2.0 * acos(0.0)));
}
double quad_area(double x1, double y1, double x2, double y2, double x3,
                 double y3, double x4, double y4) {
  return 0.5 * abs(x1 * y2 + x2 * y3 + x3 * y4 + x4 * y1 - x2 * y1 - x3 * y2 -
                   x4 * y3 - x1 * y4);
}
double cosangle(double a, double b, double c) {
  return acos((double)(b * b + c * c - a * a) / (double)(2 * b * c));
}
long long int Set(long long int N, long long int pos) {
  return N = N | (1ll << pos);
}
long long int reset(long long int N, long long int pos) {
  return N = N & ~(1ll << pos);
}
bool check(long long int N, long long int pos) {
  return (bool)(N & (1ll << pos));
}
long long int a[100005];
vector<long long int> ans;
int main() {
  int t, cno = 0;
  scanf("%d", &t);
  ;
  while (t--) {
    ans.clear();
    long long int n;
    cin >> n;
    for (long long int i = 0; i < n; i++) {
      cin >> a[i];
    }
    sort(a, a + n);
    int i = n / 2;
    i--;
    int j = i + 1;
    if (n % 2) {
      ans.push_back(a[j]);
      j++;
    }
    for (; i >= 0; i--, j++) {
      ans.push_back(a[i]);
      ans.push_back(a[j]);
    }
    for (long long int i = 0; i < n; i++) {
      if (i) {
        cout << " ";
      }
      cout << ans[i];
    }
    cout << '\n';
  }
}

```