#include <bits/stdc++.h>
#pragma GCC optimize("Ofast")
#pragma GCC target("avx,avx2,fma")
#pragma GCC optimization("unroll-loops")
using namespace std;
const long double Pi = 3.141592653;
const long long mod = 998244353;
long long INF = 1000000000000000000;
struct mycmp {
  bool operator()(pair<int, int> a, pair<int, int> b) const {
    return (a.first < b.first) || (a.first == b.first && a.second > b.second);
  }
};
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int t = 1;
  cin >> t;
  while (t--) {
    long long n;
    cin >> n;
    vector<long long> a(n);
    for (int i = 0; i < n; i++) cin >> a[i];
    long long ans5 = 0, cnt = 0;
    for (int i = 0; i < n; i++) {
      if (a[i] % 3)
        ++cnt;
      else
        ans5 = max(ans5, a[i] / 3);
    }
    if (!cnt)
      cout << ans5 << "\n";
    else {
      long long x1 = 0, x2 = 0;
      long long ans = 0, ans1 = 0, ans2 = 0;
      for (int i = 0; i < n; i++) {
        ans = max(ans, a[i] / 3);
        long long y = a[i] % 3;
        if (y == 2)
          x1 = max(x1, y);
        else if (y == 1)
          x1 = max(x1, y);
      }
      ans += x1;
      x1 = 0;
      long long ans4 = 0, o = 0;
      for (int i = 0; i < n; i++) {
        ans4 = max(ans4, a[i] / 3);
        if (a[i] == 1) ans4 = INF, ++o;
        long long y = a[i] % 3;
        if (y == 2)
          x1 = max(x1, 1LL);
        else if (y == 1)
          x1 = max(x1, 2LL);
      }
      if (x1 == 2) {
        for (int i = 0; i < n; i++) {
          if (a[i] % 3 == 1) {
            long long h = max(a[i] - 4, 0LL);
            ans1 = max(ans1, h / 3);
          } else
            ans1 = max(ans1, a[i] / 3);
        }
        ans1 += 2;
        ans4 = INF;
      } else
        ans4 += 1, ans1 = INF;
      x1 = 0, x2 = 0;
      for (int i = 0; i < n; i++) {
        if (a[i] % 3 == 1) ++x1;
        if (a[i] % 3 == 2) ++x2;
        if (a[i] % 3 == 0)
          ans2 = max(ans2, a[i] / 3 - 1);
        else
          ans2 = max(ans2, a[i] / 3);
      }
      ans2 += 2;
      if (o) ans1 = INF, ans4 = INF;
      cout << min(ans1, min(ans2, min(ans, ans4))) << "\n";
    }
  }
}

### Prompt

Please provide a cpp coded solution 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>
#pragma GCC optimize("Ofast")
#pragma GCC target("avx,avx2,fma")
#pragma GCC optimization("unroll-loops")
using namespace std;
const long double Pi = 3.141592653;
const long long mod = 998244353;
long long INF = 1000000000000000000;
struct mycmp {
  bool operator()(pair<int, int> a, pair<int, int> b) const {
    return (a.first < b.first) || (a.first == b.first && a.second > b.second);
  }
};
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int t = 1;
  cin >> t;
  while (t--) {
    long long n;
    cin >> n;
    vector<long long> a(n);
    for (int i = 0; i < n; i++) cin >> a[i];
    long long ans5 = 0, cnt = 0;
    for (int i = 0; i < n; i++) {
      if (a[i] % 3)
        ++cnt;
      else
        ans5 = max(ans5, a[i] / 3);
    }
    if (!cnt)
      cout << ans5 << "\n";
    else {
      long long x1 = 0, x2 = 0;
      long long ans = 0, ans1 = 0, ans2 = 0;
      for (int i = 0; i < n; i++) {
        ans = max(ans, a[i] / 3);
        long long y = a[i] % 3;
        if (y == 2)
          x1 = max(x1, y);
        else if (y == 1)
          x1 = max(x1, y);
      }
      ans += x1;
      x1 = 0;
      long long ans4 = 0, o = 0;
      for (int i = 0; i < n; i++) {
        ans4 = max(ans4, a[i] / 3);
        if (a[i] == 1) ans4 = INF, ++o;
        long long y = a[i] % 3;
        if (y == 2)
          x1 = max(x1, 1LL);
        else if (y == 1)
          x1 = max(x1, 2LL);
      }
      if (x1 == 2) {
        for (int i = 0; i < n; i++) {
          if (a[i] % 3 == 1) {
            long long h = max(a[i] - 4, 0LL);
            ans1 = max(ans1, h / 3);
          } else
            ans1 = max(ans1, a[i] / 3);
        }
        ans1 += 2;
        ans4 = INF;
      } else
        ans4 += 1, ans1 = INF;
      x1 = 0, x2 = 0;
      for (int i = 0; i < n; i++) {
        if (a[i] % 3 == 1) ++x1;
        if (a[i] % 3 == 2) ++x2;
        if (a[i] % 3 == 0)
          ans2 = max(ans2, a[i] / 3 - 1);
        else
          ans2 = max(ans2, a[i] / 3);
      }
      ans2 += 2;
      if (o) ans1 = INF, ans4 = INF;
      cout << min(ans1, min(ans2, min(ans, ans4))) << "\n";
    }
  }
}

```