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,os,io
input = io.BytesIO(os.read(0, os.fstat(0).st_size)).readline
#input = sys.stdin.readline
import math
import bisect
t=int(input())
for a0 in range(t):
    n=int(input())
    l=[int(i) for i in input().split()]
    z=max(l)
    f1=0;f2=0;f3=0; 
    
    c1=z//3 
    c2=(z//3)+1 
    c3=(z//3)+1
    #print(c1,c2,c3)
    for i in range(n):
        if(l[i]%3!=0):
            f1=1 
            break
    
    for i in range(n):
        if(l[i]%3==0 or l[i]%3==1):
            continue
        else:
            f2=1 
            break 
    op=0 
    for i in range(n):
        if(l[i]==1):
            #print("yes")
            op=1
        elif(l[i]%3==0 or l[i]%3==2 or (l[i]-2)%3==2):
            continue
        else:
            #print("alkj;")
            f3=1 
            break 
    #print(f1,f2,f3)
    if(f1==0):
        print(c1)
    elif(f2==0):
        print(c2)
    elif(f3==0):
        #print("here")
        if(op==1):
            print(c2+1)
        else:
            print(c2)
    
    
    