#include <bits/stdc++.h>
using namespace std;
const int N = 2010;
char s[N];
int n, k, arr[N];
long long x;
inline bool check(int val, int o, int t) {
  if (val > o + t * 2) return false;
  return (o > 0 || (val % 2 == 0));
}
void solve() {
  scanf("%d", &n);
  for (int i = 0; i < n; i++) scanf("%d", &arr[i]);
  int ans = 1000000010;
  for (int o = 0; o < 2; o++) {
    for (int mx, cur, t = 0; t < 3; t++) {
      mx = 0;
      for (int i = 0; i < n; i++) {
        cur = max(0, arr[i] / 3 - 10);
        while (cur * 3 < arr[i]) {
          if (check(arr[i] - cur * 3, o, t)) break;
          cur++;
        }
        if (cur * 3 > arr[i]) cur = 1000000010;
        mx = max(mx, cur);
      }
      ans = min(ans, mx + o + t);
    }
  }
  printf("%d\n", ans);
}
int main() {
  int t;
  cin >> t;
  while (t--) {
    solve();
  }
  return 0;
}

### 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;
const int N = 2010;
char s[N];
int n, k, arr[N];
long long x;
inline bool check(int val, int o, int t) {
  if (val > o + t * 2) return false;
  return (o > 0 || (val % 2 == 0));
}
void solve() {
  scanf("%d", &n);
  for (int i = 0; i < n; i++) scanf("%d", &arr[i]);
  int ans = 1000000010;
  for (int o = 0; o < 2; o++) {
    for (int mx, cur, t = 0; t < 3; t++) {
      mx = 0;
      for (int i = 0; i < n; i++) {
        cur = max(0, arr[i] / 3 - 10);
        while (cur * 3 < arr[i]) {
          if (check(arr[i] - cur * 3, o, t)) break;
          cur++;
        }
        if (cur * 3 > arr[i]) cur = 1000000010;
        mx = max(mx, cur);
      }
      ans = min(ans, mx + o + t);
    }
  }
  printf("%d\n", ans);
}
int main() {
  int t;
  cin >> t;
  while (t--) {
    solve();
  }
  return 0;
}

```