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;

#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace __gnu_pbds;
template <typename T> using ordered_set = tree<T, null_type,less<T>, rb_tree_tag,tree_order_statistics_node_update>;
// orderded_set ->  find_by_order(x)<itr, x being 0-indexed>; order_of_key(x)<count, strictly less>

typedef long long       ll;
typedef long double     lld;
typedef pair<int,int>   pii;
#define loop(n)         for(ll i = 0; i < (n); ++i)
#define rep(i,zz,n)     for(ll i = zz; i < (n); ++i)
#define repi(i,zz,n)    for(ll i = zz; i >= (n); --i)
#define all(v)          v.begin(),v.end()
#define pb              push_back
#define ff              first
#define ss              second
#define sz(x)           (int)x.size()
#define nline           cout << '\n'

#define     debug(x)    cout << #x << " "; _print(x); cout << '\n';
#define     _debug(a,b) cout << #a << " [ "; array_debug(a,b); cout << "]\n";
template <class T, class V> void array_debug(T *a, V n){ rep(i,0,n) cout << a[i] << ' ';}
template <class T>          void _print(T t) {cout << t;}
template <class T, class V> void _print(pair <T, V> p) {cout << "{"; _print(p.ff); cout << ","; _print(p.ss); cout << "}";}
template <class T>          void _print(vector <T> v) {cout << "[ "; for (T i : v) {_print(i); cout << " ";} cout << "]";}
template <class T>          void _print(set <T> v) {cout << "[ "; for (T i : v) {_print(i); cout << " ";} cout << "]";}
template <class T>          void _print(multiset <T> v) {cout << "[ "; for (T i : v) {_print(i); cout << " ";} cout << "]";}
template <class T, class V> void _print(map <T, V> v) {cout << "[ "; for (auto i : v) {_print(i); cout << " ";} cout << "]";}

const lld eps = 1e-9;
const int mod = 1e9 + 7;
const ll inf = 1e18;
const int N1 = 2e5 + 10;

int code(){

    int n;
    cin >> n;

    vector<ll> a(n);
    loop(n) cin >> a[i];
    sort(all(a));
    reverse(all(a));

    ll ans = LLONG_MAX;

    rep(i, 0, 4) {
        rep(j, 0, 4) {
            rep(k, max(0ll, a[0] / 3 - 4), a[0] / 3 + 1) {
                
                bool pos = 1;
                rep(l, 0, n) {
                    ll x1 = a[l] - min(a[l] / 3, k) * 3;
                    x1 -= min(x1 / 2, j) * 2;
                    x1 -= min(x1, i);
                    
                    if (x1 > 0) {
                        pos = 0;
                        break;
                    }
                }
                
                if (pos) {
                    ans = min(ans, i + j + k);
                }
                
            }
        }
    }

    cout << ans << '\n';

    return 0;
}

int main(){
    ios::sync_with_stdio(0); cin.tie(0);
    // freopen("input.txt","r",stdin);
    // freopen("output.txt","w",stdout);
    
    int tc = 1;
    cin >> tc;

    loop(tc)  code();
    // rep(i, 1, tc+1)  cout << "Case #" << i << ": ", code();

    return 0;
}