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 CF1620D extends PrintWriter {
	CF1620D() { super(System.out); }
	Scanner sc = new Scanner(System.in);
	public static void main(String[] $) {
		CF1620D o = new CF1620D(); o.main(); o.flush();
	}

	static final int INF = 2000000000;
	int solve(int a, int m1, int m2) {
		int m3 = INF;
		for (int n1 = 0; n1 <= m1; n1++)
			for (int n2 = 0; n2 <= m2; n2++) {
				int b = n1 * 1 + n2 * 2;
				if (a >= b) {
					int c = a - b;
					if (c % 3 == 0)
						m3 = Math.min(m3, c / 3);
				}
			}
		return m3;
	}
	void main() {
		int t = sc.nextInt();
		while (t-- > 0) {
			int n = sc.nextInt();
			int[] aa = new int[n];
			for (int i = 0; i < n; i++)
				aa[i] = sc.nextInt();
			int m_ = INF;
			for (int m1 = 0; m1 < 2; m1++)
				for (int m2 = 0; m2 < 3; m2++) {
					int m3 = 0;
					for (int i = 0; i < n; i++)
						m3 = Math.max(m3, solve(aa[i], m1, m2));
					m_ = Math.min(m_, m1 + m2 + m3);
				}
			println(m_);
		}
	}
}
