#include <bits/stdc++.h>
using namespace std;
const int N = 110;
int t, n;
int a[N];
int now[20];
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cin >> t;
  while (t--) {
    cin >> n;
    for (int i = 1; i <= n; ++i) cin >> a[i];
    int ans = 1e9;
    for (int cnt1 = 0; cnt1 <= 6; ++cnt1)
      for (int cnt2 = 0; cnt2 <= 3; ++cnt2) {
        for (int i = 0; i <= 12; ++i) now[i] = 0;
        now[0] = 1;
        for (int i = 1; i <= cnt1; ++i)
          for (int j = 12; j >= 1; --j) now[j] |= now[j - 1];
        for (int i = 1; i <= cnt2; ++i)
          for (int j = 12; j >= 2; --j) now[j] |= now[j - 2];
        int maxs = 0;
        for (int i = 1; i <= n; ++i) {
          int flag = 0;
          for (int j = 12; j >= 0; --j)
            if (now[j])
              if (j <= a[i] && (a[i] - j) % 3 == 0) {
                flag = 1;
                maxs = max(maxs, (a[i] - j) / 3);
                break;
              }
          if (flag == 0) maxs = 1e9;
        }
        ans = min(ans, maxs + cnt1 + cnt2);
      }
    cout << ans << "\n";
  }
}

### 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 int N = 110;
int t, n;
int a[N];
int now[20];
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cin >> t;
  while (t--) {
    cin >> n;
    for (int i = 1; i <= n; ++i) cin >> a[i];
    int ans = 1e9;
    for (int cnt1 = 0; cnt1 <= 6; ++cnt1)
      for (int cnt2 = 0; cnt2 <= 3; ++cnt2) {
        for (int i = 0; i <= 12; ++i) now[i] = 0;
        now[0] = 1;
        for (int i = 1; i <= cnt1; ++i)
          for (int j = 12; j >= 1; --j) now[j] |= now[j - 1];
        for (int i = 1; i <= cnt2; ++i)
          for (int j = 12; j >= 2; --j) now[j] |= now[j - 2];
        int maxs = 0;
        for (int i = 1; i <= n; ++i) {
          int flag = 0;
          for (int j = 12; j >= 0; --j)
            if (now[j])
              if (j <= a[i] && (a[i] - j) % 3 == 0) {
                flag = 1;
                maxs = max(maxs, (a[i] - j) / 3);
                break;
              }
          if (flag == 0) maxs = 1e9;
        }
        ans = min(ans, maxs + cnt1 + cnt2);
      }
    cout << ans << "\n";
  }
}

```