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>
using namespace std;
int t, n, a[109];
int up(int x) { return (x + 2) / 3; }
int cal(int i, int j) {
  int num = 0;
  for (int k = 1; k <= n; k++) {
    num = max(up(a[k] - i - 2 * j), num);
  }
  return num + i + j;
}
int main() {
  cin >> t;
  while (t--) {
    cin >> n;
    for (int i = 1; i <= n; i++) cin >> a[i];
    int a1 = 0, a2 = 0, a3 = 0;
    for (int i = 1; i <= n; i++) {
      if (a[i] == 1)
        a1 = 1;
      else if (a[i] % 3 == 1)
        a3 = 1;
      else if (a[i] % 3 == 2)
        a2 = 1;
    }
    int minn = 1e9;
    if (a1 == 1) {
      minn = min(minn, cal(1, a2));
    } else {
      if (a3 == 1)
        minn = min(minn, cal(0, 2));
      else
        minn = min(minn, cal(0, 1));
    }
    cout << minn << endl;
  }
  return 0;
}
