#include <bits/stdc++.h>
using namespace std;
const int maxn = 105;
int n, a[maxn];
bool Ok(int val, int c1, int c2, int c3) {
  for (int x1 = 0; x1 <= c1; ++x1) {
    for (int x2 = 0; x2 <= c2; ++x2) {
      int res = x1 + x2 * 2;
      if (res == val) return 1;
      if (res < val && abs(res - val) % 3 == 0 && abs(res - val) / 3 <= c3)
        return 1;
    }
  }
  return 0;
}
bool Check(int c1, int c2, int c3) {
  for (int i = 1; i <= n; ++i) {
    if (!Ok(a[i], c1, c2, c3)) return 0;
  }
  return 1;
}
void Solve(int Test) {
  cin >> n;
  for (int i = 1; i <= n; ++i) cin >> a[i];
  int mx = *max_element(a + 1, a + n + 1);
  int ans = 2e9;
  for (int c1 = 0; c1 <= 3; ++c1) {
    for (int c2 = 0; c2 <= 3; ++c2) {
      int c3 = (mx - c1 - 2 * c2 + 2) / 3;
      if (Check(c1, c2, c3)) ans = min(ans, c1 + c2 + c3);
    }
  }
  cout << ans << "\n";
}
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0), cout.tie(0);
  if (fopen("CF"
            ".inp",
            "r")) {
    freopen(
        "CF"
        ".inp",
        "r", stdin);
    freopen(
        "CF"
        ".out",
        "w", stdout);
  }
  int test = 1;
  cin >> test;
  for (int i = 1; i <= test; ++i) {
    Solve(i);
  }
}

### Prompt

Please formulate 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;
const int maxn = 105;
int n, a[maxn];
bool Ok(int val, int c1, int c2, int c3) {
  for (int x1 = 0; x1 <= c1; ++x1) {
    for (int x2 = 0; x2 <= c2; ++x2) {
      int res = x1 + x2 * 2;
      if (res == val) return 1;
      if (res < val && abs(res - val) % 3 == 0 && abs(res - val) / 3 <= c3)
        return 1;
    }
  }
  return 0;
}
bool Check(int c1, int c2, int c3) {
  for (int i = 1; i <= n; ++i) {
    if (!Ok(a[i], c1, c2, c3)) return 0;
  }
  return 1;
}
void Solve(int Test) {
  cin >> n;
  for (int i = 1; i <= n; ++i) cin >> a[i];
  int mx = *max_element(a + 1, a + n + 1);
  int ans = 2e9;
  for (int c1 = 0; c1 <= 3; ++c1) {
    for (int c2 = 0; c2 <= 3; ++c2) {
      int c3 = (mx - c1 - 2 * c2 + 2) / 3;
      if (Check(c1, c2, c3)) ans = min(ans, c1 + c2 + c3);
    }
  }
  cout << ans << "\n";
}
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0), cout.tie(0);
  if (fopen("CF"
            ".inp",
            "r")) {
    freopen(
        "CF"
        ".inp",
        "r", stdin);
    freopen(
        "CF"
        ".out",
        "w", stdout);
  }
  int test = 1;
  cin >> test;
  for (int i = 1; i <= test; ++i) {
    Solve(i);
  }
}

```