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 program 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 solver():
    n = int(input())

    for _ in range(n):
        n_flavors = int(input())

        cost_flavors = list(reversed(sorted(list(map(int, input().split())))))
        final = {}
        checker = True
        for i in cost_flavors:
            if checker:
                final[3] = final.get(3, 0) + i // 3
                checker = False

            if i % 3 == 2:
                final[2] = 1
            elif i % 3 == 1:
                final[1] = 1

        if final.get(3, 0) != 0 and final.get(1, 5) != 5 and final.get(2, 5) != 5:
            if cost_flavors[0] % 3 == 0:
                final[3] = final[3] - 1
            if cost_flavors[0] % 3 == 1 and (cost_flavors[0] - 1 not in cost_flavors):
                final[3] = final[3] - 1
                final[2] = 2
                final[1] = 0
            if 1 in cost_flavors:
                final[1] = 1

        print(final.get(3, 0) + final.get(2, 0) + final.get(1, 0))

solver()