#include <bits/stdc++.h>
using namespace std;
template <typename A, typename B>
ostream &operator<<(ostream &os, const pair<A, B> &p) {
  return os << '(' << p.first << ", " << p.second << ')';
}
template <typename T_container, typename T = typename enable_if<
                                    !is_same<T_container, string>::value,
                                    typename T_container::value_type>::type>
ostream &operator<<(ostream &os, const T_container &v) {
  os << '[';
  string sep;
  for (const T &x : v) os << sep << x, sep = ", ";
  return os << ']';
}
void debug_out() { cerr << endl; }
template <typename Head, typename... Tail>
void debug_out(Head H, Tail... T) {
  cerr << ' ' << H;
  debug_out(T...);
}
void run_case() {
  int N;
  cin >> N;
  vector<int> A(N);
  for (int &a : A) cin >> a;
  int MX = *max_element(A.begin(), A.end()) / 3 + 100;
  int best = MX;
  for (int ones = 0; ones <= 2; ++ones)
    for (int twos = 0; twos <= 2; ++twos) {
      int threes = 0;
      for (int a : A) {
        int least = MX;
        for (int one = 0; one <= ones; ++one)
          for (int two = 0; two <= twos; ++two) {
            int want = a - two * 2 - one;
            if (want % 3 != 0 || want < 0) continue;
            least = min(least, want / 3);
          }
        threes = max(threes, least);
      }
      best = min(best, ones + twos + threes);
    }
  cout << best << '\n';
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int tests;
  cin >> tests;
  while (tests--) run_case();
}

### Prompt

Generate 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;
template <typename A, typename B>
ostream &operator<<(ostream &os, const pair<A, B> &p) {
  return os << '(' << p.first << ", " << p.second << ')';
}
template <typename T_container, typename T = typename enable_if<
                                    !is_same<T_container, string>::value,
                                    typename T_container::value_type>::type>
ostream &operator<<(ostream &os, const T_container &v) {
  os << '[';
  string sep;
  for (const T &x : v) os << sep << x, sep = ", ";
  return os << ']';
}
void debug_out() { cerr << endl; }
template <typename Head, typename... Tail>
void debug_out(Head H, Tail... T) {
  cerr << ' ' << H;
  debug_out(T...);
}
void run_case() {
  int N;
  cin >> N;
  vector<int> A(N);
  for (int &a : A) cin >> a;
  int MX = *max_element(A.begin(), A.end()) / 3 + 100;
  int best = MX;
  for (int ones = 0; ones <= 2; ++ones)
    for (int twos = 0; twos <= 2; ++twos) {
      int threes = 0;
      for (int a : A) {
        int least = MX;
        for (int one = 0; one <= ones; ++one)
          for (int two = 0; two <= twos; ++two) {
            int want = a - two * 2 - one;
            if (want % 3 != 0 || want < 0) continue;
            least = min(least, want / 3);
          }
        threes = max(threes, least);
      }
      best = min(best, ones + twos + threes);
    }
  cout << best << '\n';
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  int tests;
  cin >> tests;
  while (tests--) run_case();
}

```