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>
using namespace std;
 
typedef long long ll;
typedef vector <int> vi;
typedef pair<int,int> pii;
typedef vector<pair<int,int>> vpii;
typedef map <int,int> mi;
typedef set<int> si;
typedef multiset<int> msi;
 
int M = 1e9+7;
ll INF = 1e18;
int NUM = 998244353;
 
#define nl <<"\n"
#define pb push_back
#define pr make_pair
#define ci cin>>
#define co cout<<
#define space <<" "
#define precision(n) std::cout << std::setprecision(n);
#define all(v) v.begin(),v.end()
#define fi first
#define se second
 
 
ll power(ll a , ll b ,ll p){      //a^b modulo p
    ll res=1,temp=a;
    if(b<=0)
        return 1;
    while(b){
        if(b%2){
            res=(res*temp)%p;
            b--;
        }
        else{
            b/=2;
            temp=(temp*temp)%p;
        }
    }
    return res;
}
 
ll inverse(ll a, ll p){
    return power(a, p-2,p);
}
 
//--------------------------------------------------------------------------------------------------//
 
int main(){
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);
 
    // #ifndef ONLINE_JUDGE
    // freopen("input.txt", "r", stdin);
    // freopen("output.txt", "w", stdout);
    // #endif
 
    int t=1;
    ci t;
    while(t--){
        int n;
        ci n;
        int arr[n];
        for(int i=0;i<n;i++){
            ci arr[i];
        }
        sort(arr,arr+n);
        int approx3 = arr[n-1]/3,ans=M;
        for(int cnt1=0;cnt1<=2;cnt1++){
            for(int cnt2=0;cnt2<=2;cnt2++){
                if(cnt1+2*cnt2>=3)
                    continue;
                for(int cnt3 = max(approx3-1,0);cnt3<=approx3;cnt3++){
                    int temp = cnt1+cnt2+cnt3;
                    if(temp>=ans)
                        continue;
                    else{
                        // co cnt1 space<<cnt2 space<<cnt3 nl;
                        int mark[n]={};
                        for(int i=0;i<n;i++){
                            for(int j=0;j<=cnt1;j++){//count1
                                for(int k=0;k<=cnt2;k++){//count2
                                    int rem = arr[i]-j-2*k;
                                    if(rem>=0 && rem%3==0 && rem/3 <=cnt3){
                                        mark[i]=1;
                                    }
                                }
                            }
                        }
                        int fl=1;
                        for(int i=0;i<n;i++){
                            if(mark[i]==0){
                                fl=0;
                                break;
                            }
                        }
                        if(fl){
                            ans=temp;
                        }
                    }
                }
            }
        }
        co ans nl;
    }
    return 0;
}