#include <bits/stdc++.h>
using namespace std;
bool che(int k, vector<int> &a) {
  int n = a.size();
  for (int i = 0; i < 3; i++) {
    for (int j = 0; j < 3; j++) {
      int x = k - i - j;
      if (x < 0) continue;
      bool chek = 1;
      for (int p = 0; p < n; p++) {
        bool h = 0;
        for (int l = 0; l <= i; l++) {
          for (int w = 0; w <= j; w++) {
            int t = a[p] - l - 2 * w;
            if ((t - 3 * x) <= 0 and t >= 0 and (t % 3 == 0)) h = 1;
          }
        }
        if (!h) {
          chek = 0;
          break;
        }
      }
      if (chek) return 1;
    }
  }
  return 0;
}
void solve() {
  int n;
  cin >> n;
  vector<int> a(n);
  for (int i = 0; i < n; i++) cin >> a[i];
  sort(a.rbegin(), a.rend());
  int k = a[0] / 3;
  if (a[0] % 3 != 0) k++;
  int l = 1, r = a[0] + 1;
  while (l != r) {
    int m = (l + r) / 2;
    if (che(m, a))
      r = m;
    else
      l = m + 1;
  }
  cout << l << "\n";
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  int tt;
  cin >> tt;
  while (tt--) solve();
  return 0;
}

### Prompt

Construct a cpp code solution to the problem outlined:

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;
bool che(int k, vector<int> &a) {
  int n = a.size();
  for (int i = 0; i < 3; i++) {
    for (int j = 0; j < 3; j++) {
      int x = k - i - j;
      if (x < 0) continue;
      bool chek = 1;
      for (int p = 0; p < n; p++) {
        bool h = 0;
        for (int l = 0; l <= i; l++) {
          for (int w = 0; w <= j; w++) {
            int t = a[p] - l - 2 * w;
            if ((t - 3 * x) <= 0 and t >= 0 and (t % 3 == 0)) h = 1;
          }
        }
        if (!h) {
          chek = 0;
          break;
        }
      }
      if (chek) return 1;
    }
  }
  return 0;
}
void solve() {
  int n;
  cin >> n;
  vector<int> a(n);
  for (int i = 0; i < n; i++) cin >> a[i];
  sort(a.rbegin(), a.rend());
  int k = a[0] / 3;
  if (a[0] % 3 != 0) k++;
  int l = 1, r = a[0] + 1;
  while (l != r) {
    int m = (l + r) / 2;
    if (che(m, a))
      r = m;
    else
      l = m + 1;
  }
  cout << l << "\n";
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  int tt;
  cin >> tt;
  while (tt--) solve();
  return 0;
}

```