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 MyClass {
    public static void main(String args[]) {
        /*
        Easy idea but
        One stupid mistake: exist number 1 => must use 1 coin
        */
      Scanner in= new Scanner(System.in);
      int t= in.nextInt();
      while(t-->0){
          int n= in.nextInt();
            int[] a= new int[n];
            for(int i=0;i<n;i++) a[i]= in.nextInt();
            int[] div= new int[3];
            boolean existOne= false;
            for(int num: a){
                div[num%3]= Math.max(div[num%3], num);
                if(num==1) existOne= true;
            }
            int res=Integer.MAX_VALUE;
            // div[1]= 1+3
            int three= Math.max(div[0]/3,Math.max(div[1]/3,div[2]/3));
            int one=0, two=0;
            if(div[2]>0) two=1;
            if(div[1]>0) one=1;
            if(one>0 && two>0 && div[0]>0) three= Math.max(Math.max(0,div[0]/3-1),Math.max(div[1]/3,div[2]/3));
            int res1= one+two+three;
            one=0;two=0;three=0;
            //div[1]= 2+2
            
            int res2= Integer.MAX_VALUE;
            if(div[1]>=4){
                three=  Math.max(div[0]/3,Math.max(Math.max(0,div[1]-4)/3,div[2]/3));
                two=2;
                res2= one+two+three;
                if(existOne) res2++;
            }

            System.out.println(Math.min(res1,res2));
      }
    }
}