#include <bits/stdc++.h>
using namespace std;
const long long mod = 998244353;
const long long N = 200005;
const long long inf = 2000000005;
inline void solve() {
  long long n;
  cin >> n;
  vector<long long> a(n);
  for (long long i = 0; i < n; ++i) cin >> a[i];
  long long ans = inf;
  for (long long i = 0; i <= 3; ++i) {
    for (long long j = 0; j <= 3; ++j) {
      long long mx = 0;
      for (long long c = 0; c < n; ++c) {
        long long pre = inf;
        for (long long o = 0; o <= i; ++o) {
          long long cur = inf;
          for (long long t = 0; t <= j; ++t) {
            long long x = a[c] - o - 2 * t;
            if (x % 3 == 0 && x >= 0) cur = min(cur, x / 3);
          }
          pre = min(pre, cur);
        }
        mx = max(mx, pre);
      }
      ans = min(ans, mx + i + j);
    }
  }
  cout << ans << "\n";
}
signed main() {
  ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  long long t = 1;
  cin >> t;
  while (t--) solve();
  return 0;
}

### Prompt

Please provide a CPP coded solution 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;
const long long mod = 998244353;
const long long N = 200005;
const long long inf = 2000000005;
inline void solve() {
  long long n;
  cin >> n;
  vector<long long> a(n);
  for (long long i = 0; i < n; ++i) cin >> a[i];
  long long ans = inf;
  for (long long i = 0; i <= 3; ++i) {
    for (long long j = 0; j <= 3; ++j) {
      long long mx = 0;
      for (long long c = 0; c < n; ++c) {
        long long pre = inf;
        for (long long o = 0; o <= i; ++o) {
          long long cur = inf;
          for (long long t = 0; t <= j; ++t) {
            long long x = a[c] - o - 2 * t;
            if (x % 3 == 0 && x >= 0) cur = min(cur, x / 3);
          }
          pre = min(pre, cur);
        }
        mx = max(mx, pre);
      }
      ans = min(ans, mx + i + j);
    }
  }
  cout << ans << "\n";
}
signed main() {
  ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  long long t = 1;
  cin >> t;
  while (t--) solve();
  return 0;
}

```