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;
using ll = long long;
using pii = pair<int, int>;
using vi = vector<int>;
using ld = long double;
const int N = 101;
const int mod = 1e9 + 7;
int a[N];
void solve() {
  int n;
  scanf("%d", &n);
  vi maxi(6);
  for (int i = 0; i < n; i++) {
    scanf("%d", &a[i]);
    maxi[a[i] % 6] = max(a[i], maxi[a[i] % 6]);
  }
  ll ans = LLONG_MAX;
  for (int i = 0; i < n; i++) {
    vi v3 = {a[i] / 3};
    if (v3[0]) v3.push_back(v3[0] - 1);
    for (auto c3 : v3) {
      for (int j = 0; j < n; j++) {
        vi v2 = {(a[j] - 3 * min(a[j] / 3, c3)) / 2};
        if (v2[0]) {
          v2.push_back(v2[0] - 1);
        }
        for (auto c2 : v2) {
          int mxm = 0;
          for (auto m : maxi) {
            m = m - 3 * min(m / 3, c3);
            m = m - 2 * min(m / 2, c2);
            mxm = max(mxm, m);
          }
          ans = min(ans, 0ll + c2 + c3 + mxm);
        }
      }
    }
  }
  printf("%lld\n", ans);
}
int main() {
  int t;
  scanf("%d", &t);
  while (t--) solve();
  return 0;
}
