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 sys
#sys.stdin = open("input.txt", "r")

for _ in range(int(input())):
    n = int(input())
    a = list(map(int, input().split()))
    a.sort(key=lambda x: -x)
    m = a[0]
    x3 = m//3
    need1, need2 = False, False
    x2 = 1 if m%3==2 else 0
    x1 = 1 if m%3==1 else 0
    for ai in a:
        if ai==m:
            continue
        else:
            if ai%3==2:
                need2 = True
            if ai%3==1:
                need1 = True
    if x1==x2==0:
        if need1==True and need2==True:
            x1,x2,x3 = 1,1,x3-1
        elif need1==True:
            x1 = 1
        elif need2==True:
            x2 = 1
    elif x1==1 and x2==0:
        if need2==True and need1==True:
            x2 += 1
        elif need2==True and 3 not in a:
            x1,x2,x3=0,2,x3-1
        elif need2==True:
            x1,x2,x3=1,1,x3
    else:
        if need1==True:
            x1+=1
    print(x1+x2+x3)