#include <bits/stdc++.h>
using namespace std;
template <class A, class B>
ostream& operator<<(ostream& out, const pair<A, B>& p) {
  return out << "(" << p.x << ", " << p.y << ")";
}
template <class A>
ostream& operator<<(ostream& out, const vector<A>& v) {
  for (int i = int(0); i < int(int((v).size())); i++) {
    if (i) out << " ";
    out << v[i];
  }
  return out;
}
int n;
vector<int> a;
inline bool read() {
  if (!(cin >> n)) return false;
  a.resize(n);
  for (int i = int(0); i < int(n); i++) cin >> a[i];
  return true;
}
bool p(int val, int c1, int c2, int c3) {
  for (int cur1 = int(0); cur1 < int(c1 + 1); cur1++)
    for (int cur2 = int(0); cur2 < int(c2 + 1); cur2++) {
      if (cur1 + 2 * cur2 > val) continue;
      if ((val - cur1 - 2 * cur2) % 3 != 0) continue;
      if ((val - cur1 - 2 * cur2) / 3 <= c3) return true;
    }
  return false;
}
bool possible(int c1, int c2, int c3) {
  for (int v : a) {
    if (!p(v, c1, c2, c3)) return false;
  }
  return true;
}
inline void solve() {
  int m = *max_element(a.begin(), a.end());
  int ans = int(1e9);
  const int MAG = 3;
  for (int c1 = int(0); c1 < int(MAG); c1++)
    for (int c2 = int(0); c2 < int(MAG); c2++) {
      int c3 = max(0, (m - c1 - 2 * c2 + 2) / 3);
      assert(c1 + 2 * c2 + 3 * c3 >= m);
      if (possible(c1, c2, c3)) ans = min(ans, c1 + c2 + c3);
    }
  cout << ans << endl;
}
int main() {
  int t;
  cin >> t;
  while (t--) {
    read();
    solve();
  }
  return 0;
}

### Prompt

Develop a solution in CPP to the problem described below:

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 A, class B>
ostream& operator<<(ostream& out, const pair<A, B>& p) {
  return out << "(" << p.x << ", " << p.y << ")";
}
template <class A>
ostream& operator<<(ostream& out, const vector<A>& v) {
  for (int i = int(0); i < int(int((v).size())); i++) {
    if (i) out << " ";
    out << v[i];
  }
  return out;
}
int n;
vector<int> a;
inline bool read() {
  if (!(cin >> n)) return false;
  a.resize(n);
  for (int i = int(0); i < int(n); i++) cin >> a[i];
  return true;
}
bool p(int val, int c1, int c2, int c3) {
  for (int cur1 = int(0); cur1 < int(c1 + 1); cur1++)
    for (int cur2 = int(0); cur2 < int(c2 + 1); cur2++) {
      if (cur1 + 2 * cur2 > val) continue;
      if ((val - cur1 - 2 * cur2) % 3 != 0) continue;
      if ((val - cur1 - 2 * cur2) / 3 <= c3) return true;
    }
  return false;
}
bool possible(int c1, int c2, int c3) {
  for (int v : a) {
    if (!p(v, c1, c2, c3)) return false;
  }
  return true;
}
inline void solve() {
  int m = *max_element(a.begin(), a.end());
  int ans = int(1e9);
  const int MAG = 3;
  for (int c1 = int(0); c1 < int(MAG); c1++)
    for (int c2 = int(0); c2 < int(MAG); c2++) {
      int c3 = max(0, (m - c1 - 2 * c2 + 2) / 3);
      assert(c1 + 2 * c2 + 3 * c3 >= m);
      if (possible(c1, c2, c3)) ans = min(ans, c1 + c2 + c3);
    }
  cout << ans << endl;
}
int main() {
  int t;
  cin >> t;
  while (t--) {
    read();
    solve();
  }
  return 0;
}

```