#include <bits/stdc++.h>
using namespace std;
const int N = 1e5 + 5;
const int mod = 1e9 + 7;
int T, n;
vector<int> v;
map<int, bool> m;
bool check(int a, int b, int c) {
  for (int i = 0; i < v.size(); i++) {
    int val = v[i];
    m.clear();
    for (int A = 0; A <= a; A++) {
      for (int B = 0; B <= b; B++) {
        for (int C = max(val / 3 - 2, 0); C <= min(val / 3, c); C++) {
          m[A + B * 2 + C * 3] = true;
        }
      }
    }
    if (!m[val]) return false;
  }
  return true;
}
void solve() {
  cin >> n;
  v.clear();
  m.clear();
  for (int i = 0; i < n; i++) {
    int x;
    cin >> x;
    if (!m[x]) v.push_back(x);
    m[x] = true;
  }
  int ans = 1e9;
  n = v.size();
  sort(v.begin(), v.end());
  for (int a = 0; a <= 2; a++) {
    for (int b = 0; b <= 2; b++) {
      for (int c = max(v.back() / 3 - 2, 0); c <= v.back() / 3; c++) {
        if (check(a, b, c)) {
          ans = min(ans, a + b + c);
        }
      }
    }
  }
  cout << ans << "\n";
  return;
}
signed main() {
  std::ios::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  ;
  cin >> T;
  while (T--) solve();
  return 0;
}

### Prompt

Your challenge is to write 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 N = 1e5 + 5;
const int mod = 1e9 + 7;
int T, n;
vector<int> v;
map<int, bool> m;
bool check(int a, int b, int c) {
  for (int i = 0; i < v.size(); i++) {
    int val = v[i];
    m.clear();
    for (int A = 0; A <= a; A++) {
      for (int B = 0; B <= b; B++) {
        for (int C = max(val / 3 - 2, 0); C <= min(val / 3, c); C++) {
          m[A + B * 2 + C * 3] = true;
        }
      }
    }
    if (!m[val]) return false;
  }
  return true;
}
void solve() {
  cin >> n;
  v.clear();
  m.clear();
  for (int i = 0; i < n; i++) {
    int x;
    cin >> x;
    if (!m[x]) v.push_back(x);
    m[x] = true;
  }
  int ans = 1e9;
  n = v.size();
  sort(v.begin(), v.end());
  for (int a = 0; a <= 2; a++) {
    for (int b = 0; b <= 2; b++) {
      for (int c = max(v.back() / 3 - 2, 0); c <= v.back() / 3; c++) {
        if (check(a, b, c)) {
          ans = min(ans, a + b + c);
        }
      }
    }
  }
  cout << ans << "\n";
  return;
}
signed main() {
  std::ios::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  ;
  cin >> T;
  while (T--) solve();
  return 0;
}

```