#include <bits/stdc++.h>
#pragma GCC optimize("-O3")
const long long N = 100005;
using namespace std;
void pairsort(long long a[], long long b[], long long n) {
  pair<long long, long long> pairt[n];
  for (long long i = 0; i < n; i++) {
    pairt[i].first = a[i];
    pairt[i].second = b[i];
  }
  sort(pairt, pairt + n);
  for (long long i = 0; i < n; i++) {
    a[i] = pairt[i].first;
    b[i] = pairt[i].second;
  }
}
bool isPrime(long long n) {
  if (n < 2) return 0;
  for (long long i = 2; i <= sqrt(n); i++) {
    if (n % i == 0) return 0;
  }
  return 1;
}
long long factorial(long long n) {
  long long i = 1;
  for (long long j = 1; j <= n; j++) i *= j;
  return i;
}
long long nCr(long long n, long long r) {
  return factorial(n) / (factorial(r) * factorial(n - r));
}
long double modd(long double x) {
  if (x >= 0) return x;
  return -1 * x;
}
long long dp(long long y) {
  long long s = 0;
  while (y != 0) {
    long long r = y % 10;
    s += r;
    y /= 10;
  }
  if (s % 10 == 0) return 1;
  return 0;
}
long long sd(long long n) {
  long long s = 0;
  while (n != 0) {
    s += n % 10;
    n /= 10;
  }
  return s;
}
long long mod = 998244353;
long long modexpo(long long x, unsigned long long y) {
  long long res = 1;
  x = x % mod;
  while (y > 0) {
    if (y & 1) res = (res * x) % mod;
    y = y >> 1;
    x = (x * x) % mod;
  }
  return res;
}
long long ps(long long n) {
  long double x = (long double)n;
  if (ceil(sqrt(x)) == floor(sqrt(x))) return 1;
  return 0;
}
vector<long long> adj[N];
vector<long long> cost;
long long src, des, w, u, v;
vector<long long> vis;
vector<long long> dist, res, parent;
long long n, k;
long long p = 1;
void dfs(long long x) {
  vis[x] = 1;
  for (long long i = 0; i < adj[x].size(); i++) {
    if (!vis[adj[x][i]]) {
      dfs(adj[x][i]);
    }
  }
}
void bfs() {
  long long s = 1;
  queue<long long> q;
  q.push(s);
  parent[s] = 1;
  vis[s] = 1;
  while (!q.empty()) {
    s = q.front();
    q.pop();
    for (long long i = 0; i < adj[s].size(); i++) {
      if (!vis[adj[s][i]]) {
        q.push(adj[s][i]);
        vis[adj[s][i]] = 1;
        parent[adj[s][i]] = s;
      }
    }
  }
}
long long g, h;
bool prime[1000005];
vector<long long> pp;
void sieve() {
  memset(prime, true, sizeof(prime));
  long long n = 1000005;
  for (long long p = 2; p * p <= n; p++) {
    if (prime[p] == true) {
      for (long long i = p * p; i <= n; i += p) prime[i] = false;
    }
  }
}
long long mod1(string num, long long a) {
  long long res = 0;
  for (long long i = 0; i < num.length(); i++)
    res = (res * 10 + (long long)num[i] - '0') % a;
  return res;
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  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 b[n];
    long long j = n - 1;
    long long i = 0, k = n - 1;
    while (i < k) b[j] = a[i], j--, b[j] = a[k], j--, k--, i++;
    if (i == k) b[j] = a[i];
    for (long long i = 0; i < n; i++) {
      cout << b[i] << " ";
    }
    cout << '\n';
  }
}

### Prompt

Create a solution in CPP for 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>
#pragma GCC optimize("-O3")
const long long N = 100005;
using namespace std;
void pairsort(long long a[], long long b[], long long n) {
  pair<long long, long long> pairt[n];
  for (long long i = 0; i < n; i++) {
    pairt[i].first = a[i];
    pairt[i].second = b[i];
  }
  sort(pairt, pairt + n);
  for (long long i = 0; i < n; i++) {
    a[i] = pairt[i].first;
    b[i] = pairt[i].second;
  }
}
bool isPrime(long long n) {
  if (n < 2) return 0;
  for (long long i = 2; i <= sqrt(n); i++) {
    if (n % i == 0) return 0;
  }
  return 1;
}
long long factorial(long long n) {
  long long i = 1;
  for (long long j = 1; j <= n; j++) i *= j;
  return i;
}
long long nCr(long long n, long long r) {
  return factorial(n) / (factorial(r) * factorial(n - r));
}
long double modd(long double x) {
  if (x >= 0) return x;
  return -1 * x;
}
long long dp(long long y) {
  long long s = 0;
  while (y != 0) {
    long long r = y % 10;
    s += r;
    y /= 10;
  }
  if (s % 10 == 0) return 1;
  return 0;
}
long long sd(long long n) {
  long long s = 0;
  while (n != 0) {
    s += n % 10;
    n /= 10;
  }
  return s;
}
long long mod = 998244353;
long long modexpo(long long x, unsigned long long y) {
  long long res = 1;
  x = x % mod;
  while (y > 0) {
    if (y & 1) res = (res * x) % mod;
    y = y >> 1;
    x = (x * x) % mod;
  }
  return res;
}
long long ps(long long n) {
  long double x = (long double)n;
  if (ceil(sqrt(x)) == floor(sqrt(x))) return 1;
  return 0;
}
vector<long long> adj[N];
vector<long long> cost;
long long src, des, w, u, v;
vector<long long> vis;
vector<long long> dist, res, parent;
long long n, k;
long long p = 1;
void dfs(long long x) {
  vis[x] = 1;
  for (long long i = 0; i < adj[x].size(); i++) {
    if (!vis[adj[x][i]]) {
      dfs(adj[x][i]);
    }
  }
}
void bfs() {
  long long s = 1;
  queue<long long> q;
  q.push(s);
  parent[s] = 1;
  vis[s] = 1;
  while (!q.empty()) {
    s = q.front();
    q.pop();
    for (long long i = 0; i < adj[s].size(); i++) {
      if (!vis[adj[s][i]]) {
        q.push(adj[s][i]);
        vis[adj[s][i]] = 1;
        parent[adj[s][i]] = s;
      }
    }
  }
}
long long g, h;
bool prime[1000005];
vector<long long> pp;
void sieve() {
  memset(prime, true, sizeof(prime));
  long long n = 1000005;
  for (long long p = 2; p * p <= n; p++) {
    if (prime[p] == true) {
      for (long long i = p * p; i <= n; i += p) prime[i] = false;
    }
  }
}
long long mod1(string num, long long a) {
  long long res = 0;
  for (long long i = 0; i < num.length(); i++)
    res = (res * 10 + (long long)num[i] - '0') % a;
  return res;
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  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 b[n];
    long long j = n - 1;
    long long i = 0, k = n - 1;
    while (i < k) b[j] = a[i], j--, b[j] = a[k], j--, k--, i++;
    if (i == k) b[j] = a[i];
    for (long long i = 0; i < n; i++) {
      cout << b[i] << " ";
    }
    cout << '\n';
  }
}

```