#include <bits/stdc++.h>
using namespace std;
const int inf = 0x3f3f3f3f;
const int maxn = 1e5 + 5;
int a[10], q[maxn];
int main() {
  int t;
  cin >> t;
  while (t--) {
    memset(a, 0, sizeof(a));
    int m1 = 0, p1 = 0, i, j, n, f = 0, ma = 0;
    cin >> n;
    for (i = 1; i <= n; i++) {
      int x;
      cin >> x;
      q[i] = x;
      if (x > 3) {
        m1 = max(m1, x / 3);
        a[x % 3] = 1;
      } else {
        a[x] = 1;
      }
      if (x == 1) f = 1;
      ma = max(ma, x);
    }
    int ans = 1e9;
    if (!a[1] && !a[2]) {
      ans = ma / 3;
    } else if (!a[1] && a[2]) {
      ans = ma / 3 + 1;
    } else if (a[1] && !a[2]) {
      ans = ma / 3 + 1;
    } else {
      int js = 0;
      for (i = 1; i <= n; i++) {
        if (q[i] % 3 == 1) {
          js = max(js, (q[i] - 1) / 3);
        } else if (q[i] % 3 == 2) {
          js = max(js, (q[i] - 2) / 3);
        } else {
          js = max(js, (q[i]) / 3 - 1);
        }
      }
      ans = min(ans, js + 2);
      if (!f) {
        js = 0;
        for (i = 1; i <= n; i++) {
          if (q[i] % 3 == 1) {
            js = max(js, (q[i] - 4) / 3);
          } else if (q[i] % 3 == 2) {
            js = max(js, (q[i] - 2) / 3);
          } else {
            js = max(js, (q[i]) / 3);
          }
        }
        ans = min(ans, js + 2);
      }
    }
    cout << ans << endl;
  }
  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 inf = 0x3f3f3f3f;
const int maxn = 1e5 + 5;
int a[10], q[maxn];
int main() {
  int t;
  cin >> t;
  while (t--) {
    memset(a, 0, sizeof(a));
    int m1 = 0, p1 = 0, i, j, n, f = 0, ma = 0;
    cin >> n;
    for (i = 1; i <= n; i++) {
      int x;
      cin >> x;
      q[i] = x;
      if (x > 3) {
        m1 = max(m1, x / 3);
        a[x % 3] = 1;
      } else {
        a[x] = 1;
      }
      if (x == 1) f = 1;
      ma = max(ma, x);
    }
    int ans = 1e9;
    if (!a[1] && !a[2]) {
      ans = ma / 3;
    } else if (!a[1] && a[2]) {
      ans = ma / 3 + 1;
    } else if (a[1] && !a[2]) {
      ans = ma / 3 + 1;
    } else {
      int js = 0;
      for (i = 1; i <= n; i++) {
        if (q[i] % 3 == 1) {
          js = max(js, (q[i] - 1) / 3);
        } else if (q[i] % 3 == 2) {
          js = max(js, (q[i] - 2) / 3);
        } else {
          js = max(js, (q[i]) / 3 - 1);
        }
      }
      ans = min(ans, js + 2);
      if (!f) {
        js = 0;
        for (i = 1; i <= n; i++) {
          if (q[i] % 3 == 1) {
            js = max(js, (q[i] - 4) / 3);
          } else if (q[i] % 3 == 2) {
            js = max(js, (q[i] - 2) / 3);
          } else {
            js = max(js, (q[i]) / 3);
          }
        }
        ans = min(ans, js + 2);
      }
    }
    cout << ans << endl;
  }
  return 0;
}

```