#include <bits/stdc++.h>
using namespace std;
int main() {
  int T;
  cin >> T;
  while (T--) {
    int n, x, ans = 1e9;
    int a[100];
    cin >> n;
    for (int i = 0; i < n; i++) cin >> a[i];
    for (int a1 = 0; a1 <= 1; a1++) {
      for (int a2 = 0; a2 <= 2; a2++) {
        int a3 = 0;
        for (int i = 0; i < n; i++) {
          if (a[i] % 3 == 0) {
            if (a1 && a2)
              a3 = max(a3, a[i] / 3 - 1);
            else
              a3 = max(a3, a[i] / 3);
          } else if (a[i] % 3 == 1) {
            if (a2 == 2 && a[i] >= 4)
              a3 = max(a3, a[i] / 3 - 1);
            else if (a1 == 1)
              a3 = max(a3, a[i] / 3);
            else {
              a3 = 1e9;
              break;
            }
          } else if (a[i] % 3 == 2) {
            if (a2)
              a3 = max(a3, a[i] / 3);
            else {
              a3 = 1e9;
              break;
            }
          }
        }
        ans = min(ans, a1 + a2 + a3);
      }
    }
    cout << ans << endl;
  }
  return 0;
}

### Prompt

Your task is to create a Cpp solution to the following problem:

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.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
  int T;
  cin >> T;
  while (T--) {
    int n, x, ans = 1e9;
    int a[100];
    cin >> n;
    for (int i = 0; i < n; i++) cin >> a[i];
    for (int a1 = 0; a1 <= 1; a1++) {
      for (int a2 = 0; a2 <= 2; a2++) {
        int a3 = 0;
        for (int i = 0; i < n; i++) {
          if (a[i] % 3 == 0) {
            if (a1 && a2)
              a3 = max(a3, a[i] / 3 - 1);
            else
              a3 = max(a3, a[i] / 3);
          } else if (a[i] % 3 == 1) {
            if (a2 == 2 && a[i] >= 4)
              a3 = max(a3, a[i] / 3 - 1);
            else if (a1 == 1)
              a3 = max(a3, a[i] / 3);
            else {
              a3 = 1e9;
              break;
            }
          } else if (a[i] % 3 == 2) {
            if (a2)
              a3 = max(a3, a[i] / 3);
            else {
              a3 = 1e9;
              break;
            }
          }
        }
        ans = min(ans, a1 + a2 + a3);
      }
    }
    cout << ans << endl;
  }
  return 0;
}

```