#include <bits/stdc++.h>
using namespace std;
#pragma GCC optimize("O3")
#pragma GCC target("sse4")
long double pi = 3.141592653589793238462643383279;
const long long INF = 1e18 + 10;
template <typename Arg1>
void __f(const char* name, Arg1&& arg1) {
  std::cerr << name << " : " << arg1 << endl;
}
template <typename Arg1, typename... Args>
void __f(const char* names, Arg1&& arg1, Args&&... args) {
  const char* comma = strchr(names + 1, ',');
  std::cerr.write(names, comma - names) << " : " << arg1 << " | ";
  __f(comma + 1, args...);
}
const long long mod = 998244353;
long long power(long long x, long long y) {
  long long res = 1;
  if (x == 0) return 0;
  while (y > 0) {
    if (y & 1) res = (res * x) % mod;
    y = y >> 1;
    x = (x * x) % mod;
  }
  return res;
}
void bruteforce() {}
void init() {}
const long long mxN = 1e5 + 10;
void solve() {
  long long n;
  cin >> n;
  long long a[n];
  for (long long i = 0; i < n; i++) {
    cin >> a[i];
  }
  bool one = 0;
  long long x = 0, y = 0, z = 0;
  for (long long i = 0; i < n; i++) {
    if (a[i] == 1) one = 1;
    if (a[i] % 3 == 0) {
      x = max(x, a[i]);
    } else if (a[i] % 3 == 1) {
      y = max(y, a[i]);
    } else
      z = max(z, a[i]);
  }
  long long val1 = max({x / 3 - ((y && z) ? 1ll : 0ll), y / 3, z / 3}) +
                   (z ? 1ll : 0ll) + (y ? 1ll : 0ll);
  long long val2 = max({x / 3, y / 3, z / 3}) + 2;
  long long val3 = max({x / 3, (y / 3 - (!one ? 1ll : 0ll)), z / 3}) + 2;
  cout << min({val1, val2, val3}) << "\n";
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  init();
  long long T = 1;
  cin >> T;
  for (long long i = 1; i <= T; i++) {
    solve();
  }
}

### Prompt

Your task is to create 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("sse4")
long double pi = 3.141592653589793238462643383279;
const long long INF = 1e18 + 10;
template <typename Arg1>
void __f(const char* name, Arg1&& arg1) {
  std::cerr << name << " : " << arg1 << endl;
}
template <typename Arg1, typename... Args>
void __f(const char* names, Arg1&& arg1, Args&&... args) {
  const char* comma = strchr(names + 1, ',');
  std::cerr.write(names, comma - names) << " : " << arg1 << " | ";
  __f(comma + 1, args...);
}
const long long mod = 998244353;
long long power(long long x, long long y) {
  long long res = 1;
  if (x == 0) return 0;
  while (y > 0) {
    if (y & 1) res = (res * x) % mod;
    y = y >> 1;
    x = (x * x) % mod;
  }
  return res;
}
void bruteforce() {}
void init() {}
const long long mxN = 1e5 + 10;
void solve() {
  long long n;
  cin >> n;
  long long a[n];
  for (long long i = 0; i < n; i++) {
    cin >> a[i];
  }
  bool one = 0;
  long long x = 0, y = 0, z = 0;
  for (long long i = 0; i < n; i++) {
    if (a[i] == 1) one = 1;
    if (a[i] % 3 == 0) {
      x = max(x, a[i]);
    } else if (a[i] % 3 == 1) {
      y = max(y, a[i]);
    } else
      z = max(z, a[i]);
  }
  long long val1 = max({x / 3 - ((y && z) ? 1ll : 0ll), y / 3, z / 3}) +
                   (z ? 1ll : 0ll) + (y ? 1ll : 0ll);
  long long val2 = max({x / 3, y / 3, z / 3}) + 2;
  long long val3 = max({x / 3, (y / 3 - (!one ? 1ll : 0ll)), z / 3}) + 2;
  cout << min({val1, val2, val3}) << "\n";
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  init();
  long long T = 1;
  cin >> T;
  for (long long i = 1; i <= T; i++) {
    solve();
  }
}

```