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.
1
1
6006

2002

import java.util.*;

public class Main {
    public static void main(String[] args) throws Exception {
        final Scanner cin = new Scanner(System.in);
        final StringBuilder cout = new StringBuilder();

        int TC = cin.nextInt();
        for (; TC-- > 0; ) {
            int n = cin.nextInt();
            List<Integer> arr = new ArrayList<>(n);
            for (int i = 0; i < n; i++)
                arr.add(cin.nextInt());
            int ans = Integer.MAX_VALUE, tans = 0, three = Collections.max(arr) / 3;
            for (int one = 0; one <= 1; one++) { // 1 can occur max once
                for (int two = 0; two <= 2; two++) {// 2 can occur max twice
                    for (int tc = three; tc >= 0 && tc >= three - 3; tc--) {
                        for (int no : arr) { // try to see if everything can be formed w/ above coins fixed
                            tans = Integer.MAX_VALUE;
                            if (no % 3 == 0) {
                                if (tc >= no / 3)
                                    tans = one + two + tc;
                                if (one > 0 && two > 0 && (1 + tc) >= no / 3)
                                    tans = one + two + tc;
                            } else if (no % 3 == 1) {
                                if (tc >= no / 3 && one > 0)
                                    tans = one + two + tc;
                                if (tc >= (no - 4) / 3 && two == 2)
                                    tans = one + two + tc;
                            } else if (no % 3 == 2) {
                                if (tc >= no / 3 && two > 0)
                                    tans = one + two + tc;
                                if (one > 0 && two == 2 && (1 + tc) >= no / 3)
                                    tans = one + two + tc;
                            }
                            if (tans == Integer.MAX_VALUE)
                                break;
                        }
                        ans = Math.min(tans, ans);
                    }
                }
            }
            cout.append(ans).append('\n');
        }
        System.out.println(cout);
    }

}
