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.
1
1
6006

2002

#include <bits/stdc++.h>
using namespace std;
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long t;
  cin >> t;
  while (t--) {
    long long n;
    cin >> n;
    long long a[n + 1];
    for (long long i = 1; i <= n; i++) {
      cin >> a[i];
    }
    bool two = 0;
    bool one = 0;
    bool onepr = 0;
    long long req[n + 1];
    for (long long i = 1; i <= n; i++) {
      if (a[i] % 3 == 1) {
        one = 1;
        req[i] = a[i] / 3 + 1;
      }
      if (a[i] % 3 == 2) {
        two = 1;
        req[i] = a[i] / 3 + 1;
      }
      if (a[i] % 3 == 0) {
        req[i] = a[i] / 3;
      }
      if (a[i] == 1) {
        onepr = 1;
      }
    }
    long long ans = a[1];
    long long ind = 0;
    for (long long i = 1; i <= n; i++) {
      if (a[i] >= ans) {
        ans = a[i];
        ind = i;
      }
    }
    if (!two && !one) {
      cout << ans / 3 << endl;
      continue;
    }
    if (!onepr) {
      if (!one) {
        ans = 0;
        for (long long i = 1; i <= n; i++) {
          if (a[i] % 3 == 2) {
            a[i] -= 2;
            req[i] = a[i] / 3;
          }
          if (req[i] > ans) {
            ans = req[i];
          }
        }
        ans++;
        cout << ans << endl;
        continue;
      }
      if (!two) {
        ans = 0;
        for (long long i = 1; i <= n; i++) {
          if (a[i] % 3 == 1) {
            a[i] -= 1;
            req[i] = a[i] / 3;
          }
          if (req[i] > ans) {
            ans = req[i];
          }
        }
        ans++;
        cout << ans << endl;
        continue;
      }
      ans = 0;
      if (a[ind] % 3 == 1) {
        for (long long i = 1; i <= n; i++) {
          if (a[i] % 3 == 2) {
            a[i] -= 2;
          }
          if (a[i] % 3 == 1) {
            a[i] -= 4;
          }
          req[i] = a[i] / 3;
          if (req[i] > ans) {
            ans = req[i];
          }
        }
        ans += 2;
        cout << ans << endl;
        continue;
      }
      if (a[ind] % 3 == 0) {
        ans = 0;
        for (long long i = 1; i <= n; i++) {
          if (a[i] % 3 == 0) {
            a[i] -= 3;
          }
          if (a[i] % 3 == 2) {
            a[i] -= 2;
          }
          if (a[i] % 3 == 1) {
            a[i] -= 1;
          }
          req[i] = a[i] / 3;
          if (req[i] > ans) {
            ans = req[i];
          }
        }
        ans += 2;
        cout << ans << endl;
      }
    }
    if (onepr) {
      if (!two) {
        ans = 0;
        for (long long i = 1; i <= n; i++) {
          if (a[i] % 3 == 1) {
            a[i] -= 1;
            req[i] = a[i] / 3;
          }
          if (req[i] > ans) {
            ans = req[i];
          }
        }
        ans++;
        cout << ans << endl;
        continue;
      }
      ans = 0;
      for (long long i = 1; i <= n; i++) {
        if (a[i] % 3 == 0) {
          a[i] -= 3;
        }
        if (a[i] % 3 == 2) {
          a[i] -= 2;
        }
        if (a[i] % 3 == 1) {
          a[i] -= 1;
        }
        req[i] = a[i] / 3;
        if (req[i] > ans) {
          ans = req[i];
        }
      }
      ans += 2;
      cout << ans << endl;
    }
  }
  return 0;
}
