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.Arrays;
import java.util.Scanner;

public class D {
    public static void main(String[] args) {
        Scanner s = new Scanner(System.in);
        int lines = s.nextInt();

        s.nextLine();

        for (int i = 0; i < lines; i += 1) {
            solve2(s.nextInt(), s);
        }
    }


    public static int solve(int n, Scanner s) {
        boolean[] mod = new boolean[3];

        boolean one = false;
        boolean two = false;

        int max = 0;
        int t;

        for (int i = 0; i < n; i++) {
            t = s.nextInt();
            mod[t % 3] = true;
            max = Math.max(max, t);

            if (t == 1) one = true;
            if (t == 2) two = true;
        }

        if (one && two) return (max - 1) / 3 + 2;

        if (one) if (mod[2]) return (max - 1) / 3 + 2; else return (max + 1) / 3 + 1;

        if (two) if (mod[1]) return (max - 1) / 3 + 2; else return (max) / 3 + 1;

        if (max % 3 == 0) return max / 3 + (mod[1] || mod[2] ? 1 : 0);
        if (max % 3 == 1) return max / 3 + 1;
        if (max % 3 == 2) return max / 3 + (mod[1] ? 2 : 1);

        return -1;
    }

    public static void solve2(int n, Scanner s) {
        long[] a = new long[n];
        for (int i = 0; i < n; i++) {
            a[i] = s.nextLong();
        }

        Arrays.sort(a);

        long min = Integer.MAX_VALUE;

        for (long three = a[n - 1] / 3 - 5; three < a[n - 1] / 3 + 5; three++) {
            for (long two = 0; two < 3; two++) {
                for (long one = 0; one < 2; one++) {
                    for (int i = 0; i < n; i++) {
                        if (a[i] == 1) if (one == 0) break;
                        if (a[i] == 2) if (two == 0) break;

                        if (a[i] % 3 == 0) if (a[i] / 3 > three + (one >= 1 && two >= 1 ? 1 : 0)) break;
                        if (a[i] % 3 == 1) if ((one >= 1 || two >= 2) && (a[i] / 3 <= three + (two >= 2 ? 1 : 0))); else break;
                        if (a[i] % 3 == 2) if ((two >= 1) && (a[i] / 3 <= three + (one >= 1 && two >= 2 ? 1 : 0))); else break;

                        if (i == n - 1) min = Math.min(min, three + two + one);
                    }
                }
            }
        }

        System.out.println(min);
    }
}
