#include <bits/stdc++.h>
using namespace std;
long long inf = 1e18;
int __FAST_IO__ = []() {
  std::ios::sync_with_stdio(0);
  std::cin.tie(0);
  std::cout.tie(0);
  return 0;
}();
int main() {
  int t;
  cin >> t;
  while (t--) {
    int N;
    cin >> N;
    vector<int> v(N);
    for (int i = 0; i < (N); ++i) cin >> v[i];
    int ans = 1e9;
    for (int i = 0; i < (5); ++i)
      for (int j = 0; j < (5); ++j) {
        int tot = i + j * 2;
        vector<int> mo(tot + 1, -1);
        for (int x = 0; x < (i + 1); ++x)
          for (int y = 0; y < (j + 1); ++y) {
            mo[x + y * 2] = x + y;
          }
        int ans2 = 0;
        for (int k = 0; k < (N); ++k) {
          bool flag = false;
          for (int x = (tot); x >= 0; --x) {
            if (mo[x] >= 0 && v[k] >= x && (v[k] - x) % 3 == 0) {
              ans2 = max(ans2, (v[k] - x) / 3);
              flag = true;
              break;
            }
          }
          if (!flag) {
            ans2 = 1e9;
            break;
          }
        }
        ans = min(ans, ans2 + i + j);
      }
    printf("%d\n", ans);
  }
  return 0;
}

### Prompt

In cpp, your task is to solve 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;
long long inf = 1e18;
int __FAST_IO__ = []() {
  std::ios::sync_with_stdio(0);
  std::cin.tie(0);
  std::cout.tie(0);
  return 0;
}();
int main() {
  int t;
  cin >> t;
  while (t--) {
    int N;
    cin >> N;
    vector<int> v(N);
    for (int i = 0; i < (N); ++i) cin >> v[i];
    int ans = 1e9;
    for (int i = 0; i < (5); ++i)
      for (int j = 0; j < (5); ++j) {
        int tot = i + j * 2;
        vector<int> mo(tot + 1, -1);
        for (int x = 0; x < (i + 1); ++x)
          for (int y = 0; y < (j + 1); ++y) {
            mo[x + y * 2] = x + y;
          }
        int ans2 = 0;
        for (int k = 0; k < (N); ++k) {
          bool flag = false;
          for (int x = (tot); x >= 0; --x) {
            if (mo[x] >= 0 && v[k] >= x && (v[k] - x) % 3 == 0) {
              ans2 = max(ans2, (v[k] - x) / 3);
              flag = true;
              break;
            }
          }
          if (!flag) {
            ans2 = 1e9;
            break;
          }
        }
        ans = min(ans, ans2 + i + j);
      }
    printf("%d\n", ans);
  }
  return 0;
}

```