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

import java.io.*;

import java.util.*;
/*

*/

 
 public class D{
	static FastReader sc=null;
	
	public static void main(String[] args) {
		sc=new FastReader();
		int t=sc.nextInt();
		while(t-->0) {
			int n=sc.nextInt();
			int a[]=sc.readArray(n);
			
			int max=0;
			boolean odd=false;
			for(int e:a) {
				max=Math.max(max, e);
				odd|=(e%2==1);
			}
			
			int one=max,two=0;
			
			if(max%2==1)two=(max+1)/2;
			else two=max/2+(odd?1:0);
			
			
			int three=0;
			boolean present[]=new boolean[3];
			for(int e:a)present[e%3]=true;
			
			if(max%3==0)three=max/3+((present[1] || present[2])?1:0);
			else {
				//2 2 3 3....
				//1 3 3 3....
				three=Math.max(3, max/3+1);
			}
			
			System.out.println(Math.min(one, Math.min(two, three)));
			
		}
	}
	
	static int[] ruffleSort(int a[]) {
		ArrayList<Integer> al=new ArrayList<>();
		for(int i:a)al.add(i);
		Collections.sort(al);
		for(int i=0;i<a.length;i++)a[i]=al.get(i);
		return a;
	}
	
	static void print(int a[]) {
		for(int e:a) {
			System.out.print(e+" ");
		}
		System.out.println();
	}
	
	static class FastReader{
		
		StringTokenizer st=new StringTokenizer("");
		BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
		
		String next() {
			while(!st.hasMoreTokens()) 
				try {
					st=new StringTokenizer(br.readLine());
				}
			   catch(IOException e){
				   e.printStackTrace();
			   }
			return st.nextToken();
		}
		
		int nextInt() {
			return Integer.parseInt(next());
		}
		
		long nextLong() {
			return Long.parseLong(next());
		}
		
		int[] readArray(int n) {
			int a[]=new int[n];
			for(int i=0;i<n;i++)a[i]=sc.nextInt();
			return a;
		}
	}
	
	
}
