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>	
 
/*
#pragma optimize ("g",on)
#pragma GCC optimize("inline")
#pragma GCC optimize ("Ofast")
#pragma GCC optimize ("unroll-loops")
#pragma GCC optimize ("03")
#pragma GCC target ("sse,sse2,sse3,ssse3,sse4,popcnt,abm,avx2,mmx,fma,avx,tune=native")
*/

//01001101 01100001 01101011 01101000 01100001  01100111 01100001 01111001 

using namespace std;

#define all(a) a.begin(), a.end()                                                   
#define rall(a) a.rbegin(), a.rend()
#define ll long long
#define pb push_back
#define sz(a) a.size()
#define nl '\n'
#define popb pop_back()                                                                   
#define ld double
#define ull unsigned long long
#define ff first                                         
#define ss second  
#define fix fixed << setprecision
#define pii pair<int, int>                          
#define E exit (0)
#define int long long
 
const int inf = 1e18, N = 1e6 + 1, mod = 998244353;                    

vector<pii> dir = {{1, 0}, {0, 1}, {-1, 0}, {0, -1}};

main() {                                                         
	//freopen("cowrect.in", "r", stdin);                                                                                     
	//freopen("cowrect.out", "w", stdout);                                                                                     
	ios_base::sync_with_stdio(NULL);                                                                                        
	cin.tie(NULL);
	int t;
	cin >> t;
	while (t--) {
		int n;
		cin >> n;
		int a[n + 1];
		for (int i = 1; i <= n; i++) cin >> a[i];
		int mx = 0, ok = 0, ok1 = 0;
		for (int i = 1; i <= n; i++) {
			if (a[i] % 3 == 1) ok = 1;
			if (a[i] % 3 == 2) ok1 = 1;
			mx = max(mx, a[i]);
		}
		if (!ok && !ok1) cout << mx / 3 << nl;
		else if (ok && !ok1) cout << mx / 3 + 1 << nl;
		else if (!ok && ok1) cout << mx / 3 + 1 << nl;
		else {
			if (mx % 3 != 2) cout << mx / 3 + 1 << nl;
			else cout << mx / 3 + 2 << nl;	
		}
	}
}                                 