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>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace std;
using namespace __gnu_pbds;

template <class T> using Tree = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;

#pragma GCC optimize("Ofast")
#pragma GCC target("avx,avx2,sse,sse2,sse3,sse4,popcnt,fma")
#pragma GCC optimize("unroll-loops")

typedef long long int ll;
typedef long double ld;

#define fastio ios_base::sync_with_stdio(false); cin.tie(NULL)
#define endl '\n'
#define pb push_back
#define conts continue
#define all(a) a.begin(),a.end()
#define rall(a) a.rbegin(), a.rend()
#define yes cout << "YES" << endl
#define no cout << "NO" << endl
#define ff first
#define ss second
#define ceil2(x,y) (x+y-1) / y
#define sz(a) a.size()
#define setbits(x) __builtin_popcountll(x)
#ifndef ONLINE_JUDGE
#define debug(x) cout << #x <<" = "; print(x); cout << endl;
#else
#define debug(x)
#endif

#define rep(i,n) for(ll i = 0; i < n; ++i)
#define rep1(i,n) for(ll i = 1; i <= n; ++i)

bool iseven(ll n) {if ((n & 1) == 0) return true; return false;}

void print(ll t) {cout << t;}
void print(int t) {cout << t;}
void print(string t) {cout << t;}
void print(char t) {cout << t;}
void print(double t) {cout << t;}
void print(ld t) {cout << t;}

template <class T, class V> void print(pair <T, V> p);
template <class T> void print(vector <T> v);
template <class T> void print(set <T> v);
template <class T, class V> void print(map <T, V> v);
template <class T> void print(multiset <T> v);
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 << "]";}

void usaco(string filename){
    freopen((filename + ".in").c_str(), "r", stdin);
    freopen((filename + ".out").c_str(), "w", stdout);
}

const ll MOD = 1e9+7;
const ll maxn = 1e5 + 5;
const ll inf = 1e18;

void solve()
{
    // https://youtu.be/0wtOyCcdUWo?t=2232
    int t; cin >> t;
    while(t--){
        ll n; cin >> n;
        vector<ll> a(n);
        rep(i,n) cin >> a[i];

        ll ans = inf;

        // not optimal to take > 2 1s
        // [1,1] = 1,2
        // [1,2] = 1,2,3

        // not optimal to take > 3 2s
        // [2,2,2] = 2,4,6
        // [1,2,3] = 1,2,3,4,5,6
        
        rep(mxones, 2){
            rep(mxtwos, 3){
                ll mxthrees = 0;

                rep(i,n){
                    ll mn = inf;
                    rep(ones, mxones+1){
                        rep(twos, mxtwos+1){
                            ll tosub = (ones * 1) + (twos * 2);
                            ll temp = a[i] - tosub;
                            if(temp % 3 != 0) conts;
                            mn = min(mn, temp / 3);
                        }
                    }

                    mxthrees = max(mxthrees, mn);
                }

                ll sum = mxones + mxtwos + mxthrees;
                ans = min(ans, sum);
            }
        }

        cout << ans << endl;
    }
}

int main()
{
    fastio;
    solve();
    return 0;
}