#include <bits/stdc++.h>
#pragma GCC optimize("O3", "Ofast", "no-stack-protector", "unroll-loops", \
                     "omit-frame-pointer", "inline")
#pragma GCC option("arch=native", "tune=native", "no-zero-upper")
using namespace std;
struct st {
  long long A, B, C;
};
bool ok(st arr, long long N) {
  for (long long i = 0; i <= arr.A; i++) {
    for (long long j = 0; j <= arr.B; j++) {
      if (i * 1 + j * 2 <= N && (N - (i * 1 + j * 2)) % 3 == 0 &&
          (N - (i * 1 + j * 2)) / 3 <= arr.C) {
        return true;
      }
    }
  }
  return false;
}
int32_t main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  vector<st> arr;
  for (long long i = 0; i <= 6; i++) {
    for (long long j = 0; j <= 6; j++) {
      for (long long k = 0; k <= 6; k++) {
        if (i + j + k <= 6) arr.push_back({i, j, k});
      }
    }
  }
  long long T;
  cin >> T;
  while (T--) {
    long long N;
    cin >> N;
    vector<long long> vect;
    long long tot = 0;
    for (long long i = 0; i < N; i++) {
      long long n;
      cin >> n;
      vect.push_back(n);
      tot = max(tot, n / 3);
    }
    tot = max(0ll, tot - 4);
    long long res = 1e18;
    for (auto x : arr) {
      x.C += tot;
      long long cont = 0;
      for (long long i = 0; i < N; i++) {
        if (ok(x, vect[i])) cont++;
      }
      if (cont == N) {
        res = min(res, x.A + x.B + x.C);
      }
    }
    cout << res << '\n';
  }
  return 0;
}

### Prompt

Your task is to create 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>
#pragma GCC optimize("O3", "Ofast", "no-stack-protector", "unroll-loops", \
                     "omit-frame-pointer", "inline")
#pragma GCC option("arch=native", "tune=native", "no-zero-upper")
using namespace std;
struct st {
  long long A, B, C;
};
bool ok(st arr, long long N) {
  for (long long i = 0; i <= arr.A; i++) {
    for (long long j = 0; j <= arr.B; j++) {
      if (i * 1 + j * 2 <= N && (N - (i * 1 + j * 2)) % 3 == 0 &&
          (N - (i * 1 + j * 2)) / 3 <= arr.C) {
        return true;
      }
    }
  }
  return false;
}
int32_t main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  vector<st> arr;
  for (long long i = 0; i <= 6; i++) {
    for (long long j = 0; j <= 6; j++) {
      for (long long k = 0; k <= 6; k++) {
        if (i + j + k <= 6) arr.push_back({i, j, k});
      }
    }
  }
  long long T;
  cin >> T;
  while (T--) {
    long long N;
    cin >> N;
    vector<long long> vect;
    long long tot = 0;
    for (long long i = 0; i < N; i++) {
      long long n;
      cin >> n;
      vect.push_back(n);
      tot = max(tot, n / 3);
    }
    tot = max(0ll, tot - 4);
    long long res = 1e18;
    for (auto x : arr) {
      x.C += tot;
      long long cont = 0;
      for (long long i = 0; i < N; i++) {
        if (ok(x, vect[i])) cont++;
      }
      if (cont == N) {
        res = min(res, x.A + x.B + x.C);
      }
    }
    cout << res << '\n';
  }
  return 0;
}

```