#include <bits/stdc++.h>
using namespace std;
constexpr long long N = 2000050;
constexpr long long D = 2000010;
constexpr long double EPS = 1e-8;
ofstream fout;
ifstream fin;
int flag = 1;
long long n, m, k;
void solve() {
  cin >> n;
  vector<long long> v(n);
  vector<long long> t;
  for (auto &x : (v)) cin >> x, t.push_back(x);
  long long b = 0;
  for (auto &x : (v)) {
    b = max(b, x / 3);
    x %= 3;
  }
  int a = 0, c = 0;
  if (count(v.begin(), v.end(), 1)) a++;
  if (count(v.begin(), v.end(), 2)) c++;
  sort(t.begin(), t.end()),
      (t).resize(unique(t.begin(), t.end()) - (t).begin());
  reverse(t.begin(), t.end());
  b += a + c;
  if (t[0] % 3 == 1 && c && t.back() != 1) {
    if (t[1] <= t[0] - 2) b--;
  }
  if (t[0] % 3 == 0 && a && c) b--;
  cout << b << "\n";
}
signed main(signed argc, char *argv[]) {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cout.precision(9);
  cout.fixed;
  srand(time(0));
  fout.precision(14);
  fout.fixed;
  int tt = 1;
  cin >> tt;
  while (tt--) {
    solve();
  }
  fout.close();
  fin.close();
}

### Prompt

Please formulate 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;
constexpr long long N = 2000050;
constexpr long long D = 2000010;
constexpr long double EPS = 1e-8;
ofstream fout;
ifstream fin;
int flag = 1;
long long n, m, k;
void solve() {
  cin >> n;
  vector<long long> v(n);
  vector<long long> t;
  for (auto &x : (v)) cin >> x, t.push_back(x);
  long long b = 0;
  for (auto &x : (v)) {
    b = max(b, x / 3);
    x %= 3;
  }
  int a = 0, c = 0;
  if (count(v.begin(), v.end(), 1)) a++;
  if (count(v.begin(), v.end(), 2)) c++;
  sort(t.begin(), t.end()),
      (t).resize(unique(t.begin(), t.end()) - (t).begin());
  reverse(t.begin(), t.end());
  b += a + c;
  if (t[0] % 3 == 1 && c && t.back() != 1) {
    if (t[1] <= t[0] - 2) b--;
  }
  if (t[0] % 3 == 0 && a && c) b--;
  cout << b << "\n";
}
signed main(signed argc, char *argv[]) {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cout.precision(9);
  cout.fixed;
  srand(time(0));
  fout.precision(14);
  fout.fixed;
  int tt = 1;
  cin >> tt;
  while (tt--) {
    solve();
  }
  fout.close();
  fin.close();
}

```