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.*;
import static java.lang.Math.*;

public class D{
	static int n;
	static long[] a;
	static boolean works(long numCoins) {
		for (int i = 0; i <= 5; i++) {
			for (int j = 0; j <= 5; j++) {
				if(solve(j, i, numCoins-i-j)) return true;
			}
		}
		return false;
	}
	
	static boolean solve(long... c) {
		List<Long> poss = new ArrayList<Long>();
		for (int i = 0; i <= c[0]; i++) {
			for (int j = 0; j <= c[1]; j++) {
				poss.add((long)(i+2*j));
			}
		}
		outer: for (int i = 0; i < a.length; i++) {
			// Try to make this shit work
			for (long x: poss) {
				long temp = a[i]-x;
				if (temp%3==0) {
					if (temp/3 <= c[2]) continue outer;
				}
			}
			return false;
		}
		return true;
	}
	
	public static void main(String[] args) throws IOException {
		// br = new BufferedReader(new FileReader(".in"));
		// out = new PrintWriter(new FileWriter(".out"));
		//new Thread(null, new (), "peepee", 1<<28).start();
		int t = readInt();
		while(t-->0) {
			n = readInt();
			a = new long[n];
			// You need to replace at most 2 3s for 3 twos  and 1 three for three ones
			for (int i =0 ; i < n; i++) a[i] =readInt();
			long l = 0, r = (int)1e9+10; 
			while(l+1<r) {
				long mid = (l+r)/2;
				if (works(mid)) r = mid;
				else l =mid;
			}
			out.println(r);
		}
		
		
		out.close();
	}
	
	/* Stupid things to try if stuck:
	 * n=1, expand bsearch range
	 * brute force small patterns
	 * submit stupid intuition
	 */
	
	static BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
	static PrintWriter out = new PrintWriter(new BufferedOutputStream(System.out));
	static StringTokenizer st = new StringTokenizer("");
	static String read() throws IOException{return st.hasMoreTokens() ? st.nextToken():(st = new StringTokenizer(br.readLine())).nextToken();}
	static int readInt() throws IOException{return Integer.parseInt(read());}
	static long readLong() throws IOException{return Long.parseLong(read());}
	static double readDouble() throws IOException{return Double.parseDouble(read());}
}