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 ExactChange
{
    public static int numcoins(int n)
    {
        int coins = 0;
        if(n%3==0)
            coins = n/3;
        else if(n%3==1)
            coins = (n-1)/3 + 1;
        else
            coins = (n-2)/3 + 1;
        return coins;
    }

    public static int numcoinmulti(int a[])
    {
        int len = a.length;
        int m = max_element(a);
        int num = numcoins(m);
        int mod3[] = {0, 0, 0};
        for(int i = 0; i<len; i++)
        {
            mod3[a[i]%3]++;
        }
        if(mod3[m%3]==len)
        {
            return num;  
        }
        else
        {
            if(m%3 == 0)
            {
                if(mod3[1]!=0 && mod3[2]==0)
                    num++;
                else if(mod3[2]!=0 && mod3[1]==0)
                    num++;
                else if(mod3[1]!=0 && mod3[2]!=0)
                    num++;
            }
            else if(m%3 == 1)
            {
                if(mod3[2]!=0);
            }
            else
            {
                if(mod3[1]!=0)
                    num++;
            }
            return num;
        }
    }

    public static void main(String[] args)
    {
        Scanner sc = new Scanner(System.in);
        int t = sc.nextInt();
        int n[] = new int[t];
        int ans[] = new int[t];
        for(int i = 0; i<t; i++)
        {
            n[i] = sc.nextInt();
            int a[] = new int[n[i]];
            for(int j = 0; j<n[i]; j++)
            {
                a[j] = sc.nextInt();
            }
            ans[i] = numcoinmulti(a);
        }
        for(int i = 0; i<t; i++)
        {
            System.out.println(ans[i]);
        }
    }

    public static int max_element(int a[])
    {
        int max = a[0];
        for(int i = 0; i<a.length; i++)
        {
            if(a[i]>max)
                max = a[i];
        }
        return max;
    }

    public static int min_element(int a[])
    {
        int min = a[0];
        for(int i = 0; i<a.length; i++)
        {
            if(a[i]<min)
                min = a[i];
        }
        return min;
    }
}
