#include <bits/stdc++.h>
using namespace std;
mt19937_64 rng(chrono::steady_clock::now().time_since_epoch().count());
template <typename T>
void _do(T x) {
  cerr << x << "\n";
}
template <typename T, typename... U>
void _do(T x, U... y) {
  cerr << x << ", ";
  _do(y...);
}
const int MOD1 = 1e9 + 7;
const int MOD2 = 998244353;
const long long INF = 3e18;
const long double PI = 3.14159265358979323846;
long long gcd(long long a, long long b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
long long fpow(long long a, long long b, long long m) {
  if (!b) return 1;
  long long ans = fpow(a * a % m, b / 2, m);
  return (b % 2 ? ans * a % m : ans);
}
long long inv(long long a, long long m) { return fpow(a, m - 2, m); }
const int N = 5e5 + 1;
long long a[105];
signed main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  long long t;
  cin >> t;
  while (t--) {
    long long n;
    cin >> n;
    for (int i = 1; i <= n; i++) cin >> a[i];
    long long ans = INF;
    for (int x = 0; x < 4; x++) {
      for (int y = 0; y < 4; y++) {
        bool can = 1;
        long long mx = 0;
        for (int i = 1; i <= n; i++) {
          long long tmp = INF;
          for (int j = 0; j < x + 1; j++) {
            for (int k = 0; k < y + 1; k++) {
              long long b = a[i] - j - k * 2;
              if (b < 0 || b % 3 != 0) continue;
              tmp = min(tmp, b / 3);
            }
          }
          if (tmp == INF) {
            can = 0;
            break;
          }
          mx = max(mx, tmp);
        }
        if (can) ans = min(ans, mx + x + y);
      }
    }
    cout << ans << "\n";
  }
}

### 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;
mt19937_64 rng(chrono::steady_clock::now().time_since_epoch().count());
template <typename T>
void _do(T x) {
  cerr << x << "\n";
}
template <typename T, typename... U>
void _do(T x, U... y) {
  cerr << x << ", ";
  _do(y...);
}
const int MOD1 = 1e9 + 7;
const int MOD2 = 998244353;
const long long INF = 3e18;
const long double PI = 3.14159265358979323846;
long long gcd(long long a, long long b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
long long fpow(long long a, long long b, long long m) {
  if (!b) return 1;
  long long ans = fpow(a * a % m, b / 2, m);
  return (b % 2 ? ans * a % m : ans);
}
long long inv(long long a, long long m) { return fpow(a, m - 2, m); }
const int N = 5e5 + 1;
long long a[105];
signed main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  long long t;
  cin >> t;
  while (t--) {
    long long n;
    cin >> n;
    for (int i = 1; i <= n; i++) cin >> a[i];
    long long ans = INF;
    for (int x = 0; x < 4; x++) {
      for (int y = 0; y < 4; y++) {
        bool can = 1;
        long long mx = 0;
        for (int i = 1; i <= n; i++) {
          long long tmp = INF;
          for (int j = 0; j < x + 1; j++) {
            for (int k = 0; k < y + 1; k++) {
              long long b = a[i] - j - k * 2;
              if (b < 0 || b % 3 != 0) continue;
              tmp = min(tmp, b / 3);
            }
          }
          if (tmp == INF) {
            can = 0;
            break;
          }
          mx = max(mx, tmp);
        }
        if (can) ans = min(ans, mx + x + y);
      }
    }
    cout << ans << "\n";
  }
}

```