#include <bits/stdc++.h>
using namespace std;
template <class T>
ostream &operator<<(ostream &os, const vector<T> &p) {
  os << "[";
  for (auto &it : p) os << it << " ";
  return os << "]";
}
template <class S, class T>
ostream &operator<<(ostream &os, const pair<S, T> &p) {
  return os << "(" << p.first << "," << p.second << ")";
}
template <class T>
inline T Bit(T x, long long i) {
  return (x >> i) & 1;
}
template <class T>
bool umin(T &a, T b) {
  return a > b ? (a = b, true) : false;
}
template <class T>
bool umax(T &a, T b) {
  return a < b ? (a = b, true) : false;
}
const long long N = 2e5 + 3;
void solve(long long tc) {
  long long n;
  cin >> n;
  vector<long long> a(n);
  for (long long i = 0; i < n; i++) cin >> a[i];
  long long ans = 1e12;
  for (long long one = 0; one < 3; one++) {
    for (long long two = 0; two < 3; two++) {
      long long cur = 0;
      for (long long i = 0; i < n; i++) {
        long long need = 1e12;
        for (long long j = 0; j <= one; j++) {
          for (long long k = 0; k <= two; k++) {
            if (a[i] - j - k * 2 >= 0 && (a[i] - j - k * 2) % 3 == 0) {
              umin(need, (a[i] - j - k * 2) / 3);
            }
          }
        }
        umax(cur, need);
      }
      umin(ans, cur + one + two);
    }
  }
  {
    cout << ans << '\n';
    return;
  };
}
signed main() {
  ios_base ::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long t = 1;
  cin >> t;
  for (long long i = 1; i <= t; i++) solve(i);
  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;
template <class T>
ostream &operator<<(ostream &os, const vector<T> &p) {
  os << "[";
  for (auto &it : p) os << it << " ";
  return os << "]";
}
template <class S, class T>
ostream &operator<<(ostream &os, const pair<S, T> &p) {
  return os << "(" << p.first << "," << p.second << ")";
}
template <class T>
inline T Bit(T x, long long i) {
  return (x >> i) & 1;
}
template <class T>
bool umin(T &a, T b) {
  return a > b ? (a = b, true) : false;
}
template <class T>
bool umax(T &a, T b) {
  return a < b ? (a = b, true) : false;
}
const long long N = 2e5 + 3;
void solve(long long tc) {
  long long n;
  cin >> n;
  vector<long long> a(n);
  for (long long i = 0; i < n; i++) cin >> a[i];
  long long ans = 1e12;
  for (long long one = 0; one < 3; one++) {
    for (long long two = 0; two < 3; two++) {
      long long cur = 0;
      for (long long i = 0; i < n; i++) {
        long long need = 1e12;
        for (long long j = 0; j <= one; j++) {
          for (long long k = 0; k <= two; k++) {
            if (a[i] - j - k * 2 >= 0 && (a[i] - j - k * 2) % 3 == 0) {
              umin(need, (a[i] - j - k * 2) / 3);
            }
          }
        }
        umax(cur, need);
      }
      umin(ans, cur + one + two);
    }
  }
  {
    cout << ans << '\n';
    return;
  };
}
signed main() {
  ios_base ::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long t = 1;
  cin >> t;
  for (long long i = 1; i <= t; i++) solve(i);
  return 0;
}

```