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

//package temp;
import java.util.*;
import java.math.*;
public class D {
    static int mx = 0, n, ans;
    static int[] a;
    
    static void  calc(int k1, int k2){
        int local_ans = 0, k3=0;
        //int k3 = Math.max( (mx - k1 - k*2) / 3, 0);
        
        for (int i=0;i<n;i++){
            int z = a[i];
            int u1 = Math.min(k1,z);
            z-=u1; 
            int u2 = Math.min(k2,z/2);
            z-=u2*2;
            int u3 = z/3;
            z-=u3*3;
            
            if (z==1 && u2>0) {u2-=1;u3+=1;}
            else if (z==1 && u1>1){u1-=2;u3+=1;}
            else if (z==2 && u1>0) {u3+=1;u1-=1;}
            else if (z>0) return;
            
            k3 = Math.max(k3, u3);
        }
        
        ans = Math.min(k1+k2+k3, ans);
        
    }
    

    
    
    public static void main(String args[]){
        Scanner in = new Scanner(System.in);
        int tn = in.nextInt();
        for (int ti=0;ti<tn;ti++){
            //case
            n = in.nextInt(); 
            a = new int[n];ans=Integer.MAX_VALUE;mx=0;
            for (int i=0;i<n;i++) {
                a[i] = in.nextInt();
                if (a[i]>mx) mx = a[i];
            }
            
            for (int k1=0;k1<20;k1++)
            for (int k2=0;k2<20;k2++)
            calc(k1,k2);
            
            System.out.println(ans);
            //endOf case
        }
    }
}