#include <bits/stdc++.h>
using namespace std;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long a, b, c, d, f, gg = 0, i, j, m, n, t, mod = 1e9 + 7, x, y;
  cin >> t;
  while (t--) {
    cin >> n;
    f = 0;
    vector<long long> v;
    for (i = 0; i < n; i++) {
      cin >> a;
      v.push_back(a);
    }
    f = INT_MAX;
    for (i = 0; i < 3; i++) {
      for (j = 0; j < 3; j++) {
        d = 0;
        for (int k = 0; k < n; k++) {
          a = v[k];
          b = INT_MAX;
          for (int ii = 0; ii <= i; ii++) {
            for (int jj = 0; jj <= j; jj++) {
              long long aa = a - ii - jj * 2;
              if (aa < 0 || aa % 3)
                ;
              else {
                b = min(b, aa / 3);
              }
            }
          }
          d = max(d, b);
        }
        f = min(f, d + i + j);
      }
    }
    cout << f << '\n';
  }
  return 0;
}

### Prompt

Create a solution in CPP for 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;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long a, b, c, d, f, gg = 0, i, j, m, n, t, mod = 1e9 + 7, x, y;
  cin >> t;
  while (t--) {
    cin >> n;
    f = 0;
    vector<long long> v;
    for (i = 0; i < n; i++) {
      cin >> a;
      v.push_back(a);
    }
    f = INT_MAX;
    for (i = 0; i < 3; i++) {
      for (j = 0; j < 3; j++) {
        d = 0;
        for (int k = 0; k < n; k++) {
          a = v[k];
          b = INT_MAX;
          for (int ii = 0; ii <= i; ii++) {
            for (int jj = 0; jj <= j; jj++) {
              long long aa = a - ii - jj * 2;
              if (aa < 0 || aa % 3)
                ;
              else {
                b = min(b, aa / 3);
              }
            }
          }
          d = max(d, b);
        }
        f = min(f, d + i + j);
      }
    }
    cout << f << '\n';
  }
  return 0;
}

```