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>
int n;
vector<int> v;
int ans;
void test(int one, int two) {
  int tans = 0;
  bitset<64> cover = 1;
  loop(i, 1, one) cover |= (cover << 1);
  loop(i, 1, two) cover |= (cover << 2);
  for (auto v : ::v) {
    roop(i, one + two + two, 0) if (cover[i] && i <= v && (v - i) % 3 == 0) {
      tans = max(tans, (v - i) / 3);
      goto ok;
    }
    return;
  ok:;
  }
  ans = min(ans, tans + one + two);
}
signed main() {
  int T;
  cin >> (T);
  loop(kase, 1, T) {
    cin >> n;
    v.resize(n);
    for (auto& vv : v) cin >> vv;
    ans = INT32_MAX;
    loop(i, 0, 2) loop(j, 0, 6) test(i, j);
    cout << ans << '\n';
  }
  return 0;
}
