human
PYTHON3 solution for "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.":
assistant
#Here is Python code to print one integer — the minimum number of coins you need to buy one bag of any flavor you'll choose in exact change.
 def solve():
    N = int(input())
    A = list(map(int,input().split()))
    mA = max(A)
    sA = set(a%3 for a in A)
    sA2 = set(A)
    if len(sA)==1:
        if mA%3==0:
            print(mA//3)
        else:
            print(mA//3+1)
    elif len(sA)==2:
        if 0 in sA and 1 in sA:
            print(mA//3+1)
        elif 0 in sA and 2 in sA:
            print(mA//3+1)
        else:
            # !
            if mA%3==1:
                if not 1 in sA2:
                    print(mA//3+1)
                else:
                    print(mA//3+2)
            if mA%3==2:
                print(mA//3+2)
    else:
        if mA%3==0:
            print(mA//3+1)
        elif mA%3==1:
            if not 1 in sA2 and not mA-1 in sA2:
                print(mA//3+1)
            else:
                print(mA//3+2)
        else:
            print(mA//3+2)

T = int(input())

for i in range(T):
    solve()