#include <bits/stdc++.h>
using namespace std;
const long long int mod = 1e9 + 7;
signed main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  long long int tc;
  cin >> tc;
  for (long long int i = 1; i <= tc; i++) {
    long long int n;
    cin >> n;
    vector<long long int> a(n);
    long long int coins = mod;
    for (long long int i = 0; i < n; i++) {
      cin >> a[i];
    }
    auto check = [&](long long int one, long long int two,
                     long long int val) -> long long int {
      vector<long long int> combination;
      for (long long int i = 0; i <= one; i++) {
        for (long long int j = 0; j <= two; j++) {
          combination.emplace_back(i + 2 * j);
        }
      }
      long long int cur_coins = mod;
      for (long long int i : combination) {
        if ((val < i) or (val - i) % 3 != 0) continue;
        cur_coins = min(cur_coins, (val - i) / 3);
      }
      return cur_coins + one + two;
    };
    for (long long int one = 0; one < 3; one++) {
      for (long long int two = 0; two < 3; two++) {
        long long int cur = 0;
        for (long long int i : a) {
          cur = max(cur, check(one, two, i));
        }
        coins = min(coins, cur);
      }
    }
    cout << coins;
    cout << "\n";
  }
  return 0;
}

### Prompt

Construct a CPP code solution to the problem outlined:

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;
const long long int mod = 1e9 + 7;
signed main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  long long int tc;
  cin >> tc;
  for (long long int i = 1; i <= tc; i++) {
    long long int n;
    cin >> n;
    vector<long long int> a(n);
    long long int coins = mod;
    for (long long int i = 0; i < n; i++) {
      cin >> a[i];
    }
    auto check = [&](long long int one, long long int two,
                     long long int val) -> long long int {
      vector<long long int> combination;
      for (long long int i = 0; i <= one; i++) {
        for (long long int j = 0; j <= two; j++) {
          combination.emplace_back(i + 2 * j);
        }
      }
      long long int cur_coins = mod;
      for (long long int i : combination) {
        if ((val < i) or (val - i) % 3 != 0) continue;
        cur_coins = min(cur_coins, (val - i) / 3);
      }
      return cur_coins + one + two;
    };
    for (long long int one = 0; one < 3; one++) {
      for (long long int two = 0; two < 3; two++) {
        long long int cur = 0;
        for (long long int i : a) {
          cur = max(cur, check(one, two, i));
        }
        coins = min(coins, cur);
      }
    }
    cout << coins;
    cout << "\n";
  }
  return 0;
}

```