#include <bits/stdc++.h>
using namespace std;
void solve() {
  int n;
  cin >> n;
  vector<int> arr(n);
  for (auto x = (0) - ((0) > (n)); x != (n) - ((0) > (n));
       ((0) < (n) ? x++ : x--))
    cin >> arr[x];
  sort(arr.begin(), arr.end());
  bool zero = false;
  bool one = false;
  bool two = false;
  vector<int> val(3);
  long long ans = arr[n - 1] / 3;
  set<int> s;
  for (auto x = (0) - ((0) > (n)); x != (n) - ((0) > (n));
       ((0) < (n) ? x++ : x--)) {
    if (arr[x] % 3 == 0) {
      zero = true;
      if (s.find(arr[x]) == s.end()) {
        val[0]++;
        s.insert(arr[x]);
      }
    } else if (arr[x] % 3 == 1) {
      one = true;
      if (s.find(arr[x]) == s.end()) {
        val[1]++;
        s.insert(arr[x]);
      }
    } else {
      two = true;
      if (s.find(arr[x]) == s.end()) {
        val[2]++;
        s.insert(arr[x]);
      }
    }
  }
  if (one) ans++;
  if (two) ans++;
  if (one && two && zero) {
    int k = arr[n - 1];
    if (k % 3 == 0) {
      ans -= 1;
    } else if (k % 3 == 1) {
      if (s.find(1) == s.end()) {
        if (s.find(k - 1) == s.end()) {
          ans--;
        }
      }
    }
  } else if (zero == false && one && two) {
    int k = arr[n - 1];
    if (k % 3 == 1 && s.find(1) == s.end()) {
      ans--;
    }
  } else {
    ans = ans;
  }
  cout << ans << "\n";
}
int main() {
  ios_base ::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  int t;
  cin >> t;
  while (t-- > 0) {
    solve();
  }
}

### Prompt

Please provide a cpp coded solution to the problem described below:

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;
void solve() {
  int n;
  cin >> n;
  vector<int> arr(n);
  for (auto x = (0) - ((0) > (n)); x != (n) - ((0) > (n));
       ((0) < (n) ? x++ : x--))
    cin >> arr[x];
  sort(arr.begin(), arr.end());
  bool zero = false;
  bool one = false;
  bool two = false;
  vector<int> val(3);
  long long ans = arr[n - 1] / 3;
  set<int> s;
  for (auto x = (0) - ((0) > (n)); x != (n) - ((0) > (n));
       ((0) < (n) ? x++ : x--)) {
    if (arr[x] % 3 == 0) {
      zero = true;
      if (s.find(arr[x]) == s.end()) {
        val[0]++;
        s.insert(arr[x]);
      }
    } else if (arr[x] % 3 == 1) {
      one = true;
      if (s.find(arr[x]) == s.end()) {
        val[1]++;
        s.insert(arr[x]);
      }
    } else {
      two = true;
      if (s.find(arr[x]) == s.end()) {
        val[2]++;
        s.insert(arr[x]);
      }
    }
  }
  if (one) ans++;
  if (two) ans++;
  if (one && two && zero) {
    int k = arr[n - 1];
    if (k % 3 == 0) {
      ans -= 1;
    } else if (k % 3 == 1) {
      if (s.find(1) == s.end()) {
        if (s.find(k - 1) == s.end()) {
          ans--;
        }
      }
    }
  } else if (zero == false && one && two) {
    int k = arr[n - 1];
    if (k % 3 == 1 && s.find(1) == s.end()) {
      ans--;
    }
  } else {
    ans = ans;
  }
  cout << ans << "\n";
}
int main() {
  ios_base ::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  int t;
  cin >> t;
  while (t-- > 0) {
    solve();
  }
}

```