#include <bits/stdc++.h>
using namespace std;
const long double scale = 1e+6;
const long long mod = 1000000007;
int ispower(long long x, long long y) {
  long long res = 1;
  while (y > 0) {
    if (y & 1) res = res * x;
    y = y >> 1;
    x = x * x;
  }
  return res;
}
const long long N = 100001;
using namespace std;
long long t;
cin >> t;
while (t--) {
  long long n;
  cin >> n;
  vector<long long> a(n);
  for (long long i = 0; i < n; i++) cin >> a[i];
  sort((a).begin(), (a).end());
  long long m = a[n - 1] / 3, ans = INT_MAX;
  for (long long i = 0; i <= 1; i++) {
    for (long long j = 0; j <= 2; j++) {
      for (long long k = max(1ll * 0, m - 4); k <= m; k++) {
        long long t = i + j + k, c = 0;
        long long f = 0;
        for (auto it : a) {
          long long flag = 0;
          for (long long x = 0; x <= i; x++)
            for (long long y = 0; y <= j; y++) {
              long long s = it - x - 2 * y;
              if (s >= 0 && s % 3 == 0 && s / 3 <= k) {
                c++;
                flag = 1;
                break;
              }
            }
          if (flag == 0) {
            f = 1;
            break;
          }
        }
        if (f == 0) ans = min(ans, t);
      }
    }
  }
  cout << ans << endl;
}
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  cout << fixed << setprecision(25);
  cerr << fixed << setprecision(10);
  auto start = std::chrono::high_resolution_clock::now();
  solve();
  auto stop = std::chrono::high_resolution_clock::now();
  auto duration =
      std::chrono::duration_cast<std::chrono::nanoseconds>(stop - start);
  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;
const long double scale = 1e+6;
const long long mod = 1000000007;
int ispower(long long x, long long y) {
  long long res = 1;
  while (y > 0) {
    if (y & 1) res = res * x;
    y = y >> 1;
    x = x * x;
  }
  return res;
}
const long long N = 100001;
using namespace std;
long long t;
cin >> t;
while (t--) {
  long long n;
  cin >> n;
  vector<long long> a(n);
  for (long long i = 0; i < n; i++) cin >> a[i];
  sort((a).begin(), (a).end());
  long long m = a[n - 1] / 3, ans = INT_MAX;
  for (long long i = 0; i <= 1; i++) {
    for (long long j = 0; j <= 2; j++) {
      for (long long k = max(1ll * 0, m - 4); k <= m; k++) {
        long long t = i + j + k, c = 0;
        long long f = 0;
        for (auto it : a) {
          long long flag = 0;
          for (long long x = 0; x <= i; x++)
            for (long long y = 0; y <= j; y++) {
              long long s = it - x - 2 * y;
              if (s >= 0 && s % 3 == 0 && s / 3 <= k) {
                c++;
                flag = 1;
                break;
              }
            }
          if (flag == 0) {
            f = 1;
            break;
          }
        }
        if (f == 0) ans = min(ans, t);
      }
    }
  }
  cout << ans << endl;
}
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  cout << fixed << setprecision(25);
  cerr << fixed << setprecision(10);
  auto start = std::chrono::high_resolution_clock::now();
  solve();
  auto stop = std::chrono::high_resolution_clock::now();
  auto duration =
      std::chrono::duration_cast<std::chrono::nanoseconds>(stop - start);
  return 0;
}

```