#include <bits/stdc++.h>
const long long inf = 1e9 + 7;
const int mod = 1e9 + 7;
const int MAX_N = 5e5 + 7;
using namespace std;
int a[MAX_N];
int ans = inf, N;
bool check(int x, int y, int z, int val) {
  for (int i = 0; i <= y; i++) {
    for (int j = 0; j <= z; j++) {
      int v = val - i * 2 - j;
      if (v >= 0 && v % 3 == 0 && x >= v / 3) return true;
    }
  }
  return false;
}
void solve(int x, int y, int z) {
  if (x < 0 || y < 0 || z < 0) return;
  for (int i = 1; i <= N; i++) {
    if (!check(x, y, z, a[i])) return;
  }
  ans = min(ans, x + y + z);
}
int main() {
  int T;
  cin >> T;
  while (T--) {
    cin >> N;
    int Max = 0;
    for (int i = 1; i <= N; i++) {
      cin >> a[i];
      Max = max(Max, a[i]);
    }
    Max /= 3;
    ans = inf;
    for (int i = Max - 10; i <= Max + 3; i++) {
      for (int j = 0; j <= 10; j++) {
        for (int k = 0; k <= 10; k++) {
          solve(i, j, k);
        }
      }
    }
    cout << ans << endl;
  }
}

### 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>
const long long inf = 1e9 + 7;
const int mod = 1e9 + 7;
const int MAX_N = 5e5 + 7;
using namespace std;
int a[MAX_N];
int ans = inf, N;
bool check(int x, int y, int z, int val) {
  for (int i = 0; i <= y; i++) {
    for (int j = 0; j <= z; j++) {
      int v = val - i * 2 - j;
      if (v >= 0 && v % 3 == 0 && x >= v / 3) return true;
    }
  }
  return false;
}
void solve(int x, int y, int z) {
  if (x < 0 || y < 0 || z < 0) return;
  for (int i = 1; i <= N; i++) {
    if (!check(x, y, z, a[i])) return;
  }
  ans = min(ans, x + y + z);
}
int main() {
  int T;
  cin >> T;
  while (T--) {
    cin >> N;
    int Max = 0;
    for (int i = 1; i <= N; i++) {
      cin >> a[i];
      Max = max(Max, a[i]);
    }
    Max /= 3;
    ans = inf;
    for (int i = Max - 10; i <= Max + 3; i++) {
      for (int j = 0; j <= 10; j++) {
        for (int k = 0; k <= 10; k++) {
          solve(i, j, k);
        }
      }
    }
    cout << ans << endl;
  }
}

```