#include <bits/stdc++.h>
using namespace std;
using LL = long long;
int getCases() {
  int cases = -1;
  enum InputType : int {
    SingleInput = 1,
    MultipleInput = 2,
    NotSet = 0,
  };
  InputType inputType = MultipleInput;
  if (inputType == SingleInput) {
    cases = 1;
  } else if (inputType == MultipleInput) {
    scanf("%d", &cases);
  } else {
    assert(false && "fix the function");
  }
  return cases;
}
int n;
int ar[105];
int big;
void Input() {
  big = 0;
  scanf("%d", &n);
  for (int i = 0; i < n; ++i) {
    scanf("%d", &ar[i]);
    big = max(big, ar[i]);
  }
}
int main() {
  int cases = getCases();
  while (cases--) {
    Input();
    int ans = INT_MAX;
    for (int a = 0; a <= 3; ++a) {
      for (int b = 0; b <= 3; ++b) {
        int c = (big - a - 2 * b + 2) / 3;
        int good = true;
        for (int i = 0; i < n; ++i) {
          int r = ceil((double)ar[i] / 3.0);
          if (ar[i] % 3 == 0) {
            if (c >= r) {
            } else if (c >= r - 1 && ((a >= 1 && b >= 1) || a >= 3)) {
            } else {
              good = false;
            }
          } else if (ar[i] % 3 == 1) {
            if (c >= r - 1 && a >= 1) {
            } else if (ar[i] > 3 && c >= r - 2 && b >= 2) {
            } else {
              good = false;
            }
          } else if (ar[i] % 3 == 2) {
            if (c >= r - 1 && (a >= 2 || b >= 1)) {
            } else {
              good = false;
            }
          }
        }
        if (good) {
          ans = min(ans, a + b + c);
        }
      }
    }
    printf("%d\n", ans);
  }
}

### 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;
int getCases() {
  int cases = -1;
  enum InputType : int {
    SingleInput = 1,
    MultipleInput = 2,
    NotSet = 0,
  };
  InputType inputType = MultipleInput;
  if (inputType == SingleInput) {
    cases = 1;
  } else if (inputType == MultipleInput) {
    scanf("%d", &cases);
  } else {
    assert(false && "fix the function");
  }
  return cases;
}
int n;
int ar[105];
int big;
void Input() {
  big = 0;
  scanf("%d", &n);
  for (int i = 0; i < n; ++i) {
    scanf("%d", &ar[i]);
    big = max(big, ar[i]);
  }
}
int main() {
  int cases = getCases();
  while (cases--) {
    Input();
    int ans = INT_MAX;
    for (int a = 0; a <= 3; ++a) {
      for (int b = 0; b <= 3; ++b) {
        int c = (big - a - 2 * b + 2) / 3;
        int good = true;
        for (int i = 0; i < n; ++i) {
          int r = ceil((double)ar[i] / 3.0);
          if (ar[i] % 3 == 0) {
            if (c >= r) {
            } else if (c >= r - 1 && ((a >= 1 && b >= 1) || a >= 3)) {
            } else {
              good = false;
            }
          } else if (ar[i] % 3 == 1) {
            if (c >= r - 1 && a >= 1) {
            } else if (ar[i] > 3 && c >= r - 2 && b >= 2) {
            } else {
              good = false;
            }
          } else if (ar[i] % 3 == 2) {
            if (c >= r - 1 && (a >= 2 || b >= 1)) {
            } else {
              good = false;
            }
          }
        }
        if (good) {
          ans = min(ans, a + b + c);
        }
      }
    }
    printf("%d\n", ans);
  }
}

```