#include <bits/stdc++.h>
using namespace std;
#pragma GCC optimize("O3")
#pragma GCC target("avx,avx2,sse,sse2,sse3,sse4,popcnt,fma")
#pragma GCC optimize("unroll-loops")
const long long int MAXN = 2e5 + 5;
const long long int mod = 1000000007;
const long long int N = 200005;
void solve() {
  int n;
  cin >> n;
  vector<long long int> vec(n);
  for (int i = 0; i < n; i++) cin >> vec[i];
  vector<long long int> ans(5, 0ll);
  for (auto x : vec) {
    if (x % 3ll) {
      ans[0] = 1e18;
      break;
    } else {
      ans[0] = max(ans[0], x / 3ll);
    }
  }
  for (auto x : vec) {
    if (x % 3ll != 2ll) {
      ans[1] = max(ans[1], x / 3ll);
    } else {
      ans[1] = 1e18;
      break;
    }
  }
  ans[1] += 1ll;
  for (auto x : vec) {
    if (x % 3ll != 1ll) {
      ans[2] = max(ans[2], x / 3ll);
    } else {
      ans[2] = 1e18;
      break;
    }
  }
  ans[2] += 1ll;
  for (auto x : vec) {
    if (x % 3ll) {
      ans[3] = max(ans[3], x / 3ll);
    } else {
      ans[3] = max(ans[3], x / 3ll - 1ll);
    }
  }
  ans[3] += 2ll;
  for (auto x : vec) {
    if (x == 1ll) {
      ans[4] = 1e18;
      break;
    }
    if (x % 3ll == 1ll)
      ans[4] = max(ans[4], (x - 4ll) / 3ll);
    else {
      ans[4] = max(ans[4], x / 3ll);
    }
  }
  ans[4] += 2ll;
  sort(ans.begin(), ans.end());
  cout << ans[0] << "\n";
  return;
}
int main() {
  ios_base ::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int t = 1;
  cin >> t;
  while (t--) {
    solve();
  }
  return 0;
}

### Prompt

Generate 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;
#pragma GCC optimize("O3")
#pragma GCC target("avx,avx2,sse,sse2,sse3,sse4,popcnt,fma")
#pragma GCC optimize("unroll-loops")
const long long int MAXN = 2e5 + 5;
const long long int mod = 1000000007;
const long long int N = 200005;
void solve() {
  int n;
  cin >> n;
  vector<long long int> vec(n);
  for (int i = 0; i < n; i++) cin >> vec[i];
  vector<long long int> ans(5, 0ll);
  for (auto x : vec) {
    if (x % 3ll) {
      ans[0] = 1e18;
      break;
    } else {
      ans[0] = max(ans[0], x / 3ll);
    }
  }
  for (auto x : vec) {
    if (x % 3ll != 2ll) {
      ans[1] = max(ans[1], x / 3ll);
    } else {
      ans[1] = 1e18;
      break;
    }
  }
  ans[1] += 1ll;
  for (auto x : vec) {
    if (x % 3ll != 1ll) {
      ans[2] = max(ans[2], x / 3ll);
    } else {
      ans[2] = 1e18;
      break;
    }
  }
  ans[2] += 1ll;
  for (auto x : vec) {
    if (x % 3ll) {
      ans[3] = max(ans[3], x / 3ll);
    } else {
      ans[3] = max(ans[3], x / 3ll - 1ll);
    }
  }
  ans[3] += 2ll;
  for (auto x : vec) {
    if (x == 1ll) {
      ans[4] = 1e18;
      break;
    }
    if (x % 3ll == 1ll)
      ans[4] = max(ans[4], (x - 4ll) / 3ll);
    else {
      ans[4] = max(ans[4], x / 3ll);
    }
  }
  ans[4] += 2ll;
  sort(ans.begin(), ans.end());
  cout << ans[0] << "\n";
  return;
}
int main() {
  ios_base ::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int t = 1;
  cin >> t;
  while (t--) {
    solve();
  }
  return 0;
}

```