#include <bits/stdc++.h>
using namespace std;
const int maxn = 1e2 + 10;
long long n, a[maxn], ans;
long long res(long long one, long long two) {
  long long ans = 0;
  for (int i = 0; i < n; i++) {
    long long kom = a[i], lom = INT64_MAX - 100;
    for (int j = 0; j <= one; j++) {
      for (int z = 0; z <= two; z++) {
        if (0 <= (kom - j - (2 * z)) && (kom - j - (2 * z)) % 3 == 0) {
          lom = min(lom, (kom - j - (2 * z)) / 3);
        }
      }
    }
    ans = max(ans, lom);
  }
  return ans;
}
void solve() {
  cin >> n;
  ans = INT64_MAX;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
  }
  for (int i = 0; i <= 5; i++) {
    for (int j = 0; j <= 5; j++) {
      ans = min(ans, res(i, j) + i + j);
    }
  }
  cout << ans << '\n';
}
int main() {
  ios_base ::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  int tc = 1;
  cin >> tc;
  while (tc--) {
    solve();
  }
  return 0;
}

### Prompt

Please formulate 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;
const int maxn = 1e2 + 10;
long long n, a[maxn], ans;
long long res(long long one, long long two) {
  long long ans = 0;
  for (int i = 0; i < n; i++) {
    long long kom = a[i], lom = INT64_MAX - 100;
    for (int j = 0; j <= one; j++) {
      for (int z = 0; z <= two; z++) {
        if (0 <= (kom - j - (2 * z)) && (kom - j - (2 * z)) % 3 == 0) {
          lom = min(lom, (kom - j - (2 * z)) / 3);
        }
      }
    }
    ans = max(ans, lom);
  }
  return ans;
}
void solve() {
  cin >> n;
  ans = INT64_MAX;
  for (int i = 0; i < n; i++) {
    cin >> a[i];
  }
  for (int i = 0; i <= 5; i++) {
    for (int j = 0; j <= 5; j++) {
      ans = min(ans, res(i, j) + i + j);
    }
  }
  cout << ans << '\n';
}
int main() {
  ios_base ::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  int tc = 1;
  cin >> tc;
  while (tc--) {
    solve();
  }
  return 0;
}

```