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 main() {
  int tests;
  cin >> tests;
  for (int idk = 0; idk < tests; idk++) {
    long long n;
    cin >> n;
    long long a[n], one = 0, md1 = 0, md2 = 0;
    for (int i = 0; i < n; i++) {
      cin >> a[i];
      if (a[i] == 1) one = 1;
      if (a[i] % 3 == 1) md1 = 1;
      if (a[i] % 3 == 2) md2 = 1;
    }
    std::sort(a, a + n);
    long long ans[4];
    ans[1] = 0;
    ans[2] = 0;
    if (one) ans[1] = 1;
    if (md2) ans[2] = 1;
    if ((md1) && (one == 0) && (md2)) ans[2] = 2;
    if ((md1) && (md2 == 0)) ans[1] = 1;
    long long dif = ans[1] + ans[2] * 2;
    ans[3] = 0;
    for (int i = 0; i < n; i++) {
      if ((dif == 4) && (a[n - 1] - a[i] < 3) && (a[i] % 3 == 0)) {
        ans[2]++;
        dif = 6;
      }
      ans[3] = max(ans[3], (a[i] + 2 - dif) / 3);
    }
    cout << ans[1] + ans[2] + ans[3] << endl;
  }
}
