#include <bits/stdc++.h>
using namespace std;
long long int gcd(long long int a, long long int b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long int t;
  cin >> t;
  while (t--) {
    long long int n, i, j, count = 0;
    cin >> n;
    long long int arr[n + 1];
    long long int mx = 0;
    bool a = 0, b = 0;
    set<long long int> s;
    set<long long int> s1;
    for (i = 1; i <= n; i++) {
      cin >> arr[i];
      mx = max(arr[i], mx);
      s1.insert(arr[i]);
      if (arr[i] % 3 == 1) {
        s.insert(arr[i]);
        a = 1;
      } else if (arr[i] % 3 == 2)
        b = 1;
    }
    if (mx % 3 == 0) {
      if (a == 1 || b == 1)
        cout << mx / 3 + 1 << '\n';
      else
        cout << mx / 3 << '\n';
    } else {
      if (mx < 3)
        cout << a + b << '\n';
      else if (mx % 3 == 1) {
        if ((s1.find(mx - 1) != s1.end() && b == 1) ||
            (b == 1 && s1.find(1) != s1.end()))
          cout << mx / 3 + 2 << '\n';
        else
          cout << mx / 3 + 1 << '\n';
      } else if (a == 1 && b == 1)
        cout << mx / 3 + 2 << '\n';
      else
        cout << mx / 3 + 1 << '\n';
    }
  }
  return 0;
}

### Prompt

Your challenge is to write a CPP solution to 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;
long long int gcd(long long int a, long long int b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long int t;
  cin >> t;
  while (t--) {
    long long int n, i, j, count = 0;
    cin >> n;
    long long int arr[n + 1];
    long long int mx = 0;
    bool a = 0, b = 0;
    set<long long int> s;
    set<long long int> s1;
    for (i = 1; i <= n; i++) {
      cin >> arr[i];
      mx = max(arr[i], mx);
      s1.insert(arr[i]);
      if (arr[i] % 3 == 1) {
        s.insert(arr[i]);
        a = 1;
      } else if (arr[i] % 3 == 2)
        b = 1;
    }
    if (mx % 3 == 0) {
      if (a == 1 || b == 1)
        cout << mx / 3 + 1 << '\n';
      else
        cout << mx / 3 << '\n';
    } else {
      if (mx < 3)
        cout << a + b << '\n';
      else if (mx % 3 == 1) {
        if ((s1.find(mx - 1) != s1.end() && b == 1) ||
            (b == 1 && s1.find(1) != s1.end()))
          cout << mx / 3 + 2 << '\n';
        else
          cout << mx / 3 + 1 << '\n';
      } else if (a == 1 && b == 1)
        cout << mx / 3 + 2 << '\n';
      else
        cout << mx / 3 + 1 << '\n';
    }
  }
  return 0;
}

```