#include <bits/stdc++.h>
using namespace std;
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  long long t;
  cin >> t;
  while (t--) {
    long long n;
    cin >> n;
    vector<long long> a(n);
    long long mx = 1;
    long long mn = 1000000000;
    long long second = 0;
    vector<long long> freq(3, 0);
    for (long long& ai : a) {
      cin >> ai;
      if (ai > mx) {
        second = mx;
        mx = ai;
      }
      if (ai != mx) {
        second = max(second, ai);
      }
      mn = min(mn, ai);
      freq[ai % 3]++;
    }
    long long ans;
    if (mx % 3 == 0) {
      if (freq[1] == 0 && freq[2] == 0) {
        ans = mx / 3;
      } else {
        ans = mx / 3 + 1;
      }
    } else if (mx % 3 == 2) {
      if (freq[1] == 0) {
        ans = mx / 3 + 1;
      } else {
        ans = mx / 3 + 2;
      }
    } else {
      if (freq[2] == 0 || (mn != 1 && second != mx - 1)) {
        ans = mx / 3 + 1;
      } else {
        ans = mx / 3 + 2;
      }
    }
    cout << ans << '\n';
  }
  return 0;
}

### Prompt

Develop a solution in cpp 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;
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  long long t;
  cin >> t;
  while (t--) {
    long long n;
    cin >> n;
    vector<long long> a(n);
    long long mx = 1;
    long long mn = 1000000000;
    long long second = 0;
    vector<long long> freq(3, 0);
    for (long long& ai : a) {
      cin >> ai;
      if (ai > mx) {
        second = mx;
        mx = ai;
      }
      if (ai != mx) {
        second = max(second, ai);
      }
      mn = min(mn, ai);
      freq[ai % 3]++;
    }
    long long ans;
    if (mx % 3 == 0) {
      if (freq[1] == 0 && freq[2] == 0) {
        ans = mx / 3;
      } else {
        ans = mx / 3 + 1;
      }
    } else if (mx % 3 == 2) {
      if (freq[1] == 0) {
        ans = mx / 3 + 1;
      } else {
        ans = mx / 3 + 2;
      }
    } else {
      if (freq[2] == 0 || (mn != 1 && second != mx - 1)) {
        ans = mx / 3 + 1;
      } else {
        ans = mx / 3 + 2;
      }
    }
    cout << ans << '\n';
  }
  return 0;
}

```