#include <bits/stdc++.h>
using namespace std;
const int MOD = 1000000007;
void solve() {
  int n;
  cin >> n;
  vector<int> arr(n);
  for (int i = 0; i < n; ++i) cin >> arr[i];
  sort(arr.begin(), arr.end());
  set<int> s;
  for (int num : arr) s.insert(num);
  if (n == 1)
    cout << arr[0] / 3 + (arr[0] % 3 == 0 ? 0 : 1) << endl;
  else {
    if (arr[0] == 1) {
      if (arr[1] == 2) {
        if (arr[n - 1] % 3 == 0)
          cout << arr[n - 1] / 3 + 1 << endl;
        else
          cout << arr[n - 1] / 3 + 2 << endl;
      } else {
        bool two = false;
        for (int num : arr)
          if (num % 3 == 2) {
            two = true;
            break;
          }
        if (two && arr[n - 1] % 3 == 0)
          cout << arr[n - 1] / 3 + 1 << endl;
        else if (two)
          cout << arr[n - 1] / 3 + 2 << endl;
        else
          cout << arr[n - 1] / 3 + 1 << endl;
      }
    } else {
      set<int> mods;
      for (int num : arr)
        if (num % 3) mods.insert(num % 3);
      if (mods.size() == 0)
        cout << arr[n - 1] / 3 << endl;
      else if (mods.size() == 1)
        cout << arr[n - 1] / 3 + 1 << endl;
      else if ((!s.count(arr[n - 1] / 3 * 3) && arr[n - 1] % 3 == 1) ||
               arr[n - 1] % 3 == 0)
        cout << arr[n - 1] / 3 + 1 << endl;
      else
        cout << arr[n - 1] / 3 + 2 << endl;
    }
  }
}
int main() {
  int t;
  cin >> t;
  for (int i = 0; i < t; ++i) solve();
  return 0;
}

### Prompt

Create a solution in cpp for the following problem:

One day, early in the morning, you decided to buy yourself a bag of chips in the nearby store. The store has chips of n different flavors. A bag of the i-th flavor costs a_i burles.

The store may run out of some flavors, so you'll decide which one to buy after arriving there. But there are two major flaws in this plan: 

  1. you have only coins of 1, 2 and 3 burles; 
  2. since it's morning, the store will ask you to pay in exact change, i. e. if you choose the i-th flavor, you'll have to pay exactly a_i burles. 



Coins are heavy, so you'd like to take the least possible number of coins in total. That's why you are wondering: what is the minimum total number of coins you should take with you, so you can buy a bag of chips of any flavor in exact change?

Input

The first line contains a single integer t (1 ≤ t ≤ 1000) — the number of test cases.

The first line of each test case contains the single integer n (1 ≤ n ≤ 100) — the number of flavors in the store.

The second line of each test case contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9) — the cost of one bag of each flavor.

Output

For each test case, print one integer — the minimum number of coins you need to buy one bag of any flavor you'll choose in exact change.

Example

Input


4
1
1337
3
10 8 10
5
1 2 3 4 5
3
7 77 777


Output


446
4
3
260

Note

In the first test case, you should, for example, take with you 445 coins of value 3 and 1 coin of value 2. So, 1337 = 445 ⋅ 3 + 1 ⋅ 2.

In the second test case, you should, for example, take 2 coins of value 3 and 2 coins of value 2. So you can pay either exactly 8 = 2 ⋅ 3 + 1 ⋅ 2 or 10 = 2 ⋅ 3 + 2 ⋅ 2.

In the third test case, it's enough to take 1 coin of value 3 and 2 coins of value 1.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
const int MOD = 1000000007;
void solve() {
  int n;
  cin >> n;
  vector<int> arr(n);
  for (int i = 0; i < n; ++i) cin >> arr[i];
  sort(arr.begin(), arr.end());
  set<int> s;
  for (int num : arr) s.insert(num);
  if (n == 1)
    cout << arr[0] / 3 + (arr[0] % 3 == 0 ? 0 : 1) << endl;
  else {
    if (arr[0] == 1) {
      if (arr[1] == 2) {
        if (arr[n - 1] % 3 == 0)
          cout << arr[n - 1] / 3 + 1 << endl;
        else
          cout << arr[n - 1] / 3 + 2 << endl;
      } else {
        bool two = false;
        for (int num : arr)
          if (num % 3 == 2) {
            two = true;
            break;
          }
        if (two && arr[n - 1] % 3 == 0)
          cout << arr[n - 1] / 3 + 1 << endl;
        else if (two)
          cout << arr[n - 1] / 3 + 2 << endl;
        else
          cout << arr[n - 1] / 3 + 1 << endl;
      }
    } else {
      set<int> mods;
      for (int num : arr)
        if (num % 3) mods.insert(num % 3);
      if (mods.size() == 0)
        cout << arr[n - 1] / 3 << endl;
      else if (mods.size() == 1)
        cout << arr[n - 1] / 3 + 1 << endl;
      else if ((!s.count(arr[n - 1] / 3 * 3) && arr[n - 1] % 3 == 1) ||
               arr[n - 1] % 3 == 0)
        cout << arr[n - 1] / 3 + 1 << endl;
      else
        cout << arr[n - 1] / 3 + 2 << endl;
    }
  }
}
int main() {
  int t;
  cin >> t;
  for (int i = 0; i < t; ++i) solve();
  return 0;
}

```