#include <bits/stdc++.h>
using namespace std;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  int t;
  cin >> t;
  while (t--) {
    int n;
    cin >> n;
    vector<int> arr(n);
    for (int &e : arr) cin >> e;
    int ans = *max_element(arr.begin(), arr.end());
    int ans2 = INT_MAX;
    int cond = 1;
    for (int i : arr) {
      if (i & 1) {
        cond = 0;
        break;
      }
    }
    if (cond) ans2 = ans / 2;
    int ans3 = INT_MAX;
    cond = 1;
    for (int i : arr) {
      if (i % 3 != 0) {
        cond = 0;
        break;
      }
    }
    if (cond) ans3 = ans / 3;
    int c1 = 0, c2 = 0, c3 = 0;
    int ans4 = INT_MIN;
    for (int i : arr) {
      c2 = max(c2, i / 2);
      c1 = max(c1, i - 2 * (i / 2));
    }
    ans4 = c1 + c2;
    c1 = 0, c2 = 0, c3 = 0;
    int ans5 = INT_MIN;
    for (int i : arr) {
      c3 = max(c3, i / 3);
      c1 = max(c1, i - 3 * (i / 3));
    }
    ans5 = c1 + c3;
    c1 = 0, c2 = 0, c3 = 0;
    int ans6 = INT_MIN;
    for (int i : arr) {
      if (i == 1) {
        ans6 = INT_MAX;
        break;
      }
      if (i % 3 == 0) {
        c3 = max(c3, i / 3);
      } else if ((i - 2) % 3 == 0) {
        c3 = max(c3, (i - 2) / 3);
        c1 = max(c1, (i - 3 * ((i - 2) / 3)) / 2);
      } else {
        c3 = max(c3, (i - 4) / 3);
        c1 = max(c1, (i - 3 * ((i - 4) / 3)) / 2);
      }
      ans6 = max(ans6, c1 + c3);
    }
    c1 = 0, c2 = 0, c3 = 0;
    int ans7 = INT_MIN;
    for (int i : arr) {
      c3 = max(c3, i / 3);
      int r = i - 3 * (i / 3);
      if (r == 1 and !c1) {
        c1 = 1;
      } else if (!c2) {
        c2 = 1;
      }
    }
    ans7 = c1 + c2 + c3;
    cond = 1;
    if (c1 * c2) {
      int p3 = c3 - 1;
      for (int i : arr) {
        if (i <= 3 * p3 + 3) {
          continue;
        } else {
          cond = 0;
          break;
        }
      }
    }
    if (cond and c1 * c2) ans7--;
    vector<int> v = {ans, ans2, ans3, ans4, ans5, ans6, ans7};
    cout << *min_element(v.begin(), v.end()) << endl;
  }
}

### Prompt

Your challenge is to write 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>
using namespace std;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cout.tie(nullptr);
  int t;
  cin >> t;
  while (t--) {
    int n;
    cin >> n;
    vector<int> arr(n);
    for (int &e : arr) cin >> e;
    int ans = *max_element(arr.begin(), arr.end());
    int ans2 = INT_MAX;
    int cond = 1;
    for (int i : arr) {
      if (i & 1) {
        cond = 0;
        break;
      }
    }
    if (cond) ans2 = ans / 2;
    int ans3 = INT_MAX;
    cond = 1;
    for (int i : arr) {
      if (i % 3 != 0) {
        cond = 0;
        break;
      }
    }
    if (cond) ans3 = ans / 3;
    int c1 = 0, c2 = 0, c3 = 0;
    int ans4 = INT_MIN;
    for (int i : arr) {
      c2 = max(c2, i / 2);
      c1 = max(c1, i - 2 * (i / 2));
    }
    ans4 = c1 + c2;
    c1 = 0, c2 = 0, c3 = 0;
    int ans5 = INT_MIN;
    for (int i : arr) {
      c3 = max(c3, i / 3);
      c1 = max(c1, i - 3 * (i / 3));
    }
    ans5 = c1 + c3;
    c1 = 0, c2 = 0, c3 = 0;
    int ans6 = INT_MIN;
    for (int i : arr) {
      if (i == 1) {
        ans6 = INT_MAX;
        break;
      }
      if (i % 3 == 0) {
        c3 = max(c3, i / 3);
      } else if ((i - 2) % 3 == 0) {
        c3 = max(c3, (i - 2) / 3);
        c1 = max(c1, (i - 3 * ((i - 2) / 3)) / 2);
      } else {
        c3 = max(c3, (i - 4) / 3);
        c1 = max(c1, (i - 3 * ((i - 4) / 3)) / 2);
      }
      ans6 = max(ans6, c1 + c3);
    }
    c1 = 0, c2 = 0, c3 = 0;
    int ans7 = INT_MIN;
    for (int i : arr) {
      c3 = max(c3, i / 3);
      int r = i - 3 * (i / 3);
      if (r == 1 and !c1) {
        c1 = 1;
      } else if (!c2) {
        c2 = 1;
      }
    }
    ans7 = c1 + c2 + c3;
    cond = 1;
    if (c1 * c2) {
      int p3 = c3 - 1;
      for (int i : arr) {
        if (i <= 3 * p3 + 3) {
          continue;
        } else {
          cond = 0;
          break;
        }
      }
    }
    if (cond and c1 * c2) ans7--;
    vector<int> v = {ans, ans2, ans3, ans4, ans5, ans6, ans7};
    cout << *min_element(v.begin(), v.end()) << endl;
  }
}

```