#include <bits/stdc++.h>
using namespace std;
const long long INF = 1e16;
const int N = 1e6 + 7;
int main() {
  ios::sync_with_stdio(0);
  cin.tie(nullptr);
  int tt;
  cin >> tt;
  while (tt--) {
    int n;
    cin >> n;
    bool has1 = 0;
    int mx0 = -1;
    int mx1 = -1;
    int mx2 = -1;
    for (int i = 0; i < n; i++) {
      int x;
      cin >> x;
      if (x % 3 == 0) mx0 = max(mx0, x);
      if (x % 3 == 1) mx1 = max(mx1, x);
      if (x % 3 == 2) mx2 = max(mx2, x);
      if (x == 1) has1 = 1;
    }
    if (mx1 == -1 && mx2 == -1)
      cout << mx0 / 3 << '\n';
    else {
      if (mx1 != -1 && mx2 != -1) {
        int u = max({mx1, mx2, mx0});
        if (u == mx0) cout << mx0 / 3 + 1 << '\n';
        if (u == mx2) cout << mx2 / 3 + 2 << '\n';
        if (u == mx1) {
          if (!has1 && mx0 != mx1 - 1)
            cout << mx1 / 3 + 1 << '\n';
          else
            cout << mx1 / 3 + 2 << '\n';
        }
      } else
        cout << max({mx0, mx1, mx2}) / 3 + 1 << '\n';
    }
  }
  return 0;
}

### 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;
const long long INF = 1e16;
const int N = 1e6 + 7;
int main() {
  ios::sync_with_stdio(0);
  cin.tie(nullptr);
  int tt;
  cin >> tt;
  while (tt--) {
    int n;
    cin >> n;
    bool has1 = 0;
    int mx0 = -1;
    int mx1 = -1;
    int mx2 = -1;
    for (int i = 0; i < n; i++) {
      int x;
      cin >> x;
      if (x % 3 == 0) mx0 = max(mx0, x);
      if (x % 3 == 1) mx1 = max(mx1, x);
      if (x % 3 == 2) mx2 = max(mx2, x);
      if (x == 1) has1 = 1;
    }
    if (mx1 == -1 && mx2 == -1)
      cout << mx0 / 3 << '\n';
    else {
      if (mx1 != -1 && mx2 != -1) {
        int u = max({mx1, mx2, mx0});
        if (u == mx0) cout << mx0 / 3 + 1 << '\n';
        if (u == mx2) cout << mx2 / 3 + 2 << '\n';
        if (u == mx1) {
          if (!has1 && mx0 != mx1 - 1)
            cout << mx1 / 3 + 1 << '\n';
          else
            cout << mx1 / 3 + 2 << '\n';
        }
      } else
        cout << max({mx0, mx1, mx2}) / 3 + 1 << '\n';
    }
  }
  return 0;
}

```