#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using ii = pair<int, int>;
const int MAXN = 1e5;
int solve(int ones, int twos, vector<int>& v, int n) {
  int ans = 0;
  for (int i = 0; i < n; i++) {
    if (v[i] % 3 == 0) {
      int cand = v[i] / 3;
      if (ones and twos) {
        cand--;
      }
      ans = max(ans, cand);
    } else if (v[i] % 3 == 1) {
      if (ones) {
        ans = max(ans, v[i] / 3);
      } else if (v[i] == 1) {
        return 1e9;
      } else if (twos > 1) {
        ans = max(ans, (v[i] - 4) / 3);
      } else {
        return 1e9;
      }
    } else {
      if (ones > 1) {
        ans = max(ans, v[i] / 3);
      } else if (twos) {
        ans = max(ans, v[i] / 3);
      } else {
        return 1e9;
      }
    }
  }
  return ones + twos + ans;
}
int tc() {
  int n;
  scanf("%d", &n);
  vector<int> v(n);
  for (int i = 0; i < n; i++) {
    scanf("%d", &v[i]);
  }
  sort(v.begin(), v.end());
  int ans = 1e9;
  for (int i = 0; i < 3; i++) {
    for (int j = 0; j < 3; j++) {
      int cand = solve(i, j, v, n);
      ans = min(ans, cand);
    }
  }
  return ans;
}
int main() {
  int t;
  scanf("%d", &t);
  while (t--) {
    printf("%d\n", tc());
  }
}

### Prompt

Develop a solution in cpp to the problem described below:

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;
using ll = long long;
using ii = pair<int, int>;
const int MAXN = 1e5;
int solve(int ones, int twos, vector<int>& v, int n) {
  int ans = 0;
  for (int i = 0; i < n; i++) {
    if (v[i] % 3 == 0) {
      int cand = v[i] / 3;
      if (ones and twos) {
        cand--;
      }
      ans = max(ans, cand);
    } else if (v[i] % 3 == 1) {
      if (ones) {
        ans = max(ans, v[i] / 3);
      } else if (v[i] == 1) {
        return 1e9;
      } else if (twos > 1) {
        ans = max(ans, (v[i] - 4) / 3);
      } else {
        return 1e9;
      }
    } else {
      if (ones > 1) {
        ans = max(ans, v[i] / 3);
      } else if (twos) {
        ans = max(ans, v[i] / 3);
      } else {
        return 1e9;
      }
    }
  }
  return ones + twos + ans;
}
int tc() {
  int n;
  scanf("%d", &n);
  vector<int> v(n);
  for (int i = 0; i < n; i++) {
    scanf("%d", &v[i]);
  }
  sort(v.begin(), v.end());
  int ans = 1e9;
  for (int i = 0; i < 3; i++) {
    for (int j = 0; j < 3; j++) {
      int cand = solve(i, j, v, n);
      ans = min(ans, cand);
    }
  }
  return ans;
}
int main() {
  int t;
  scanf("%d", &t);
  while (t--) {
    printf("%d\n", tc());
  }
}

```