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

#include<bits/stdc++.h>
#define int  long long
#define ull unsigned ll
#define f first
#define s second
#define pii pair<int,int>
#define pll pair<ll,ll>
#define pb push_back
#define epb emplace_back
#define inf 1e9+1
#define linf 1e18+1
using namespace std;
void solve(){
    int n;cin>>n;
    int a[n+1];
    for(int i=1;i<=n;i++)cin>>a[i];
    int ans=inf;
    for(int i=0;i<6;i++){
        for(int j=0;j<6;j++){
            int mn=-1;
            for(int k=1;k<=n;k++){
                int mn1=inf;
                for(int x=0;x<=i;x++){
                    for(int y=0;y<=j;y++){
                        if((a[k]-(x*2+y))%3==0)mn1=min(mn1,(a[k]-(x*2+y))/3);
                    }
                }
                mn=max(mn1,mn);
            }
            ans=min(ans,mn+i+j);
        }
    }
    cout<<ans<<"\n";
}

main(){
    ios_base::sync_with_stdio(false);cin.tie(0);
    int test=1;cin>>test;
    while(test--)solve();
}

