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 mincoin(arr):
    zero = 0
    num3 = max(arr)//3 - 1
    mid = num3*3+3
    lower = [0,0,0]
    upper = [0,0]
    for i in arr:
        if i < mid:
            lower[i%3] = 1
        elif i == mid:
            zero = 1
        else:
            upper[i%3-1] = 1
    lower[0] = zero
    if sum(upper) == 2:
        return num3 + 3
    elif sum(upper) == 0:
        return min(2,  sum(lower)) + num3
    elif upper[0] == 1:
        if num3 < 0:
            return sum(upper)
        elif min(arr) == 1:
            return num3 + 2 + lower[2]
        else:
            return num3 + 2 + (lower[0] and lower[2] == 1)
    else:
        if lower[1] == 1:
            return num3 + 3
        else:
            return num3 + 2

        
r = int(input())
res = []
for i in range(r):
    input()
    arr = list(map(int, input().split()))
    res.append(mincoin(arr))
    
for i in res:
    print(i)