#include <bits/stdc++.h>
using namespace std;
const long long MAX = 1e11;
const long long mod = 1e9 + 7;
const long long inf = 1000000000000000000;
void solve() {
  long long n;
  cin >> n;
  vector<long long> a(n);
  for (long long i = 0; i < n; i++) {
    cin >> a[i];
  }
  sort(a.begin(), a.end());
  long long mx = a[n - 1];
  long long c0 = 0, c1 = 0, c2 = 0;
  map<long long, long long> freq;
  for (long long i = 0; i < n; i++) {
    freq[a[i]]++;
    if (a[i] % 3 == 0) {
      c0++;
    } else if (a[i] % 3 == 1) {
      c1++;
    } else if (a[i] % 3 == 2) {
      c2++;
    }
  }
  if (mx % 3 == 0) {
    if (c1 == 0 && c2 == 0) {
      cout << mx / 3 << '\n';
      return;
    } else {
      cout << (mx / 3 - 1) + 2 << '\n';
      return;
    }
  } else if (mx % 3 == 1) {
    if (c2 == 0) {
      cout << mx / 3 + 1 << '\n';
    } else if (freq[1] == 0 && freq[mx - 1] == 0) {
      cout << (mx / 3 - 1) + 2 << '\n';
      return;
    } else {
      cout << (mx / 3 - 1) + 3 << '\n';
      return;
    }
  } else if (mx % 3 == 2) {
    if (c1 == 0) {
      cout << mx / 3 + 1 << '\n';
      return;
    } else {
      cout << mx / 3 + 2 << '\n';
      return;
    }
  }
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  ;
  long long t;
  cin >> t;
  while (t--) {
    solve();
  }
  return 0;
}

### Prompt

Please create a solution in cpp 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 MAX = 1e11;
const long long mod = 1e9 + 7;
const long long inf = 1000000000000000000;
void solve() {
  long long n;
  cin >> n;
  vector<long long> a(n);
  for (long long i = 0; i < n; i++) {
    cin >> a[i];
  }
  sort(a.begin(), a.end());
  long long mx = a[n - 1];
  long long c0 = 0, c1 = 0, c2 = 0;
  map<long long, long long> freq;
  for (long long i = 0; i < n; i++) {
    freq[a[i]]++;
    if (a[i] % 3 == 0) {
      c0++;
    } else if (a[i] % 3 == 1) {
      c1++;
    } else if (a[i] % 3 == 2) {
      c2++;
    }
  }
  if (mx % 3 == 0) {
    if (c1 == 0 && c2 == 0) {
      cout << mx / 3 << '\n';
      return;
    } else {
      cout << (mx / 3 - 1) + 2 << '\n';
      return;
    }
  } else if (mx % 3 == 1) {
    if (c2 == 0) {
      cout << mx / 3 + 1 << '\n';
    } else if (freq[1] == 0 && freq[mx - 1] == 0) {
      cout << (mx / 3 - 1) + 2 << '\n';
      return;
    } else {
      cout << (mx / 3 - 1) + 3 << '\n';
      return;
    }
  } else if (mx % 3 == 2) {
    if (c1 == 0) {
      cout << mx / 3 + 1 << '\n';
      return;
    } else {
      cout << mx / 3 + 2 << '\n';
      return;
    }
  }
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  ;
  long long t;
  cin >> t;
  while (t--) {
    solve();
  }
  return 0;
}

```