#include <bits/stdc++.h>
constexpr int N = 105;
int T, n;
int a[N];
int main() {
  std::ios::sync_with_stdio(false), std::cin.tie(nullptr);
  std::cin >> T;
  while (T--) {
    std::cin >> n;
    int max = 0;
    for (int i = 1; i <= n; ++i) {
      std::cin >> a[i];
      max = std::max(max, (a[i] + 2) / 3);
    }
    bool f1 = true;
    bool f2 = true;
    bool exist1 = false;
    bool exist2 = false;
    bool has1 = false;
    bool t1 = true;
    bool t2 = true;
    for (int i = 1; i <= n; ++i) {
      if (a[i] % 3) f1 = false;
      if (a[i] == 3 * max) f2 = false;
      if (a[i] % 3 == 1) exist1 = true;
      if (a[i] % 3 == 2) exist2 = true;
      if (a[i] == 1) has1 = true;
      if (a[i] == 3 * max - 3) t1 = false;
      if (a[i] == 3 * max - 1) t2 = false;
    }
    if (f1)
      std::cout << max << '\n';
    else if (f2) {
      if (!exist1 || !exist2)
        std::cout << max << '\n';
      else {
        if (max == 1)
          std::cout << 2 << '\n';
        else if (t1 && t2 && !has1)
          std::cout << max << '\n';
        else
          std::cout << max + 1 << '\n';
      }
    } else
      std::cout << max + 1 << '\n';
  }
  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>
constexpr int N = 105;
int T, n;
int a[N];
int main() {
  std::ios::sync_with_stdio(false), std::cin.tie(nullptr);
  std::cin >> T;
  while (T--) {
    std::cin >> n;
    int max = 0;
    for (int i = 1; i <= n; ++i) {
      std::cin >> a[i];
      max = std::max(max, (a[i] + 2) / 3);
    }
    bool f1 = true;
    bool f2 = true;
    bool exist1 = false;
    bool exist2 = false;
    bool has1 = false;
    bool t1 = true;
    bool t2 = true;
    for (int i = 1; i <= n; ++i) {
      if (a[i] % 3) f1 = false;
      if (a[i] == 3 * max) f2 = false;
      if (a[i] % 3 == 1) exist1 = true;
      if (a[i] % 3 == 2) exist2 = true;
      if (a[i] == 1) has1 = true;
      if (a[i] == 3 * max - 3) t1 = false;
      if (a[i] == 3 * max - 1) t2 = false;
    }
    if (f1)
      std::cout << max << '\n';
    else if (f2) {
      if (!exist1 || !exist2)
        std::cout << max << '\n';
      else {
        if (max == 1)
          std::cout << 2 << '\n';
        else if (t1 && t2 && !has1)
          std::cout << max << '\n';
        else
          std::cout << max + 1 << '\n';
      }
    } else
      std::cout << max + 1 << '\n';
  }
  return 0;
}

```