#include <bits/stdc++.h>
using namespace std;
int fast_mod_pow(int x, int y, int mod) {
  int res = 1;
  while (y > 0) {
    if ((y & 1) == 1) {
      res = ((long long)res * x) % mod;
    }
    y >>= 1;
    x = ((long long)x * x) % mod;
  }
  return res;
}
int inverse_mod(int n, int p) { return fast_mod_pow(n, p - 2, p); }
int divide_mod(int a, int b, int mod) {
  return ((long long)a * inverse_mod(b, mod)) % mod;
}
void solve() {
  int n;
  cin >> n;
  vector<int> A(n);
  int maxi = -1;
  for (int i = 0; i < n; ++i) {
    cin >> A[i];
    maxi = max(maxi, A[i]);
  }
  int best = 2e9;
  maxi /= 3;
  for (int ones = 0; ones <= 3; ++ones) {
    for (int twos = 0; twos <= 3; ++twos) {
      for (int threes = max(0, maxi - 3); threes <= maxi + 2; ++threes) {
        int k = ones + twos + threes;
        if (k >= best) {
          break;
        }
        bool valid = true;
        for (int i = 0; i < n; ++i) {
          int a = A[i];
          bool found = false;
          for (int x = 0; x <= ones; ++x) {
            for (int y = 0; y <= twos; ++y) {
              int rem = a - x - 2 * y;
              if (rem >= 0 and rem % 3 == 0 and rem / 3 <= threes) {
                found = true;
                break;
              }
            }
          }
          if (not found) {
            valid = false;
            break;
          }
        }
        if (valid) {
          best = k;
        }
      }
    }
  }
  cout << best << '\n';
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int t;
  cin >> t;
  while (t--) {
    solve();
  }
  return 0;
}

### Prompt

Construct a cpp code solution to the problem outlined:

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;
int fast_mod_pow(int x, int y, int mod) {
  int res = 1;
  while (y > 0) {
    if ((y & 1) == 1) {
      res = ((long long)res * x) % mod;
    }
    y >>= 1;
    x = ((long long)x * x) % mod;
  }
  return res;
}
int inverse_mod(int n, int p) { return fast_mod_pow(n, p - 2, p); }
int divide_mod(int a, int b, int mod) {
  return ((long long)a * inverse_mod(b, mod)) % mod;
}
void solve() {
  int n;
  cin >> n;
  vector<int> A(n);
  int maxi = -1;
  for (int i = 0; i < n; ++i) {
    cin >> A[i];
    maxi = max(maxi, A[i]);
  }
  int best = 2e9;
  maxi /= 3;
  for (int ones = 0; ones <= 3; ++ones) {
    for (int twos = 0; twos <= 3; ++twos) {
      for (int threes = max(0, maxi - 3); threes <= maxi + 2; ++threes) {
        int k = ones + twos + threes;
        if (k >= best) {
          break;
        }
        bool valid = true;
        for (int i = 0; i < n; ++i) {
          int a = A[i];
          bool found = false;
          for (int x = 0; x <= ones; ++x) {
            for (int y = 0; y <= twos; ++y) {
              int rem = a - x - 2 * y;
              if (rem >= 0 and rem % 3 == 0 and rem / 3 <= threes) {
                found = true;
                break;
              }
            }
          }
          if (not found) {
            valid = false;
            break;
          }
        }
        if (valid) {
          best = k;
        }
      }
    }
  }
  cout << best << '\n';
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int t;
  cin >> t;
  while (t--) {
    solve();
  }
  return 0;
}

```