#include <bits/stdc++.h>
using namespace std;
const long long MAXN = 1e5 + 123;
const long long MAX = 1000000;
const long long MOD = 1000000007;
const long double pi = 3.1415926536;
bool isPrime(long long n) {
  if (n <= 1) return 0;
  if (n <= 3) return 1;
  if (n % 2 == 0 || n % 3 == 0) return 0;
  for (auto i = 5; i * i <= n; i += 6)
    if (n % i == 0 || n % (i + 2) == 0) return 0;
  return 1;
}
long long cc(long long ans) {
  long long cc = (ans * (ans + 1)) / 2;
  return cc;
}
long long gcd(long long a, long long b) {
  if (!b) return a;
  return gcd(b, a % b);
}
void f() {
  long long n;
  cin >> n;
  long long a[n];
  for (auto i = 0; i < n; i++) cin >> a[i];
  long long c1 = 0, c2 = 0;
  sort(a, a + n);
  long long ans = a[n - 1] / 3;
  for (auto i = 0; i < n; i++) {
    if (a[i] % 3 == 1)
      ++c1;
    else if (a[i] % 3 == 2)
      ++c2;
  }
  if ((a[n - 1] % 3 == 0) && (c1 || c2)) ++ans;
  if ((a[n - 1] % 3 == 2)) {
    ans++;
    if (c1) ++ans;
  }
  if (a[n - 1] % 3 == 1) {
    ++ans;
    if (c2) {
      for (auto i = 0; i < n; i++)
        if (a[i] == 1 || (a[i] == (a[n - 1] - 1))) {
          ++ans;
          break;
        }
    }
  }
  cout << ans << "\n";
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long t;
  cin >> t;
  while (t--) {
    f();
  }
}

### Prompt

Create a solution in Cpp for 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 MAXN = 1e5 + 123;
const long long MAX = 1000000;
const long long MOD = 1000000007;
const long double pi = 3.1415926536;
bool isPrime(long long n) {
  if (n <= 1) return 0;
  if (n <= 3) return 1;
  if (n % 2 == 0 || n % 3 == 0) return 0;
  for (auto i = 5; i * i <= n; i += 6)
    if (n % i == 0 || n % (i + 2) == 0) return 0;
  return 1;
}
long long cc(long long ans) {
  long long cc = (ans * (ans + 1)) / 2;
  return cc;
}
long long gcd(long long a, long long b) {
  if (!b) return a;
  return gcd(b, a % b);
}
void f() {
  long long n;
  cin >> n;
  long long a[n];
  for (auto i = 0; i < n; i++) cin >> a[i];
  long long c1 = 0, c2 = 0;
  sort(a, a + n);
  long long ans = a[n - 1] / 3;
  for (auto i = 0; i < n; i++) {
    if (a[i] % 3 == 1)
      ++c1;
    else if (a[i] % 3 == 2)
      ++c2;
  }
  if ((a[n - 1] % 3 == 0) && (c1 || c2)) ++ans;
  if ((a[n - 1] % 3 == 2)) {
    ans++;
    if (c1) ++ans;
  }
  if (a[n - 1] % 3 == 1) {
    ++ans;
    if (c2) {
      for (auto i = 0; i < n; i++)
        if (a[i] == 1 || (a[i] == (a[n - 1] - 1))) {
          ++ans;
          break;
        }
    }
  }
  cout << ans << "\n";
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long t;
  cin >> t;
  while (t--) {
    f();
  }
}

```