#include <bits/stdc++.h>
using namespace std;
long long t;
long long n;
long long a[110];
int main(void) {
  scanf("%lld", &t);
  while (t--) {
    scanf("%lld", &n);
    for (long long i = 1; i <= n; i++) scanf("%lld", &a[i]);
    long long chk[10] = {
        0,
    };
    for (long long i = 1; i <= n; i++) chk[a[i] % 3] = 1;
    long long ans = 0;
    if (!chk[1] && !chk[2]) {
      for (long long i = 1; i <= n; i++) ans = max(a[i] / 3, ans);
    } else if (chk[1] && !chk[2]) {
      for (long long i = 1; i <= n; i++) ans = max(ans, a[i] / 3);
      ans++;
    } else if (!chk[1] && chk[2]) {
      for (long long i = 1; i <= n; i++) ans = max(ans, a[i] / 3);
      ans++;
    } else {
      long long ff = 0;
      for (long long i = 1; i <= n; i++) {
        if (a[i] == 1) ff = 1;
      }
      if (ff) {
        for (long long i = 1; i <= n; i++) {
          if (a[i] % 3 == 0)
            ans = max(ans, a[i] / 3 - 1);
          else
            ans = max(ans, a[i] / 3);
        }
        ans += 2;
      } else {
        long long hubo2 = 0;
        for (long long i = 1; i <= n; i++) {
          if (a[i] % 3 == 0)
            hubo2 = max(hubo2, a[i] / 3 - 1);
          else
            hubo2 = max(hubo2, a[i] / 3);
        }
        hubo2 += 2;
        long long hubo = 0;
        for (long long i = 1; i <= n; i++) {
          if (a[i] % 3 == 1)
            hubo = max(hubo, (a[i] - 4) / 3);
          else
            hubo = max(hubo, a[i] / 3);
        }
        hubo += 2;
        for (long long i = 1; i <= n; i++) {
          if (a[i] % 3 == 1)
            ans = max(ans, (a[i] - 4) / 3);
          else if (a[i] % 3 == 0)
            ans = max(ans, a[i] / 3 - 2);
          else
            ans = max(ans, (a[i] - 2) / 3);
        }
        ans += 3;
        ans = min(ans, hubo);
        ans = min(ans, hubo2);
      }
    }
    printf("%lld\n", ans);
  }
  return 0;
}

### Prompt

Generate 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;
long long t;
long long n;
long long a[110];
int main(void) {
  scanf("%lld", &t);
  while (t--) {
    scanf("%lld", &n);
    for (long long i = 1; i <= n; i++) scanf("%lld", &a[i]);
    long long chk[10] = {
        0,
    };
    for (long long i = 1; i <= n; i++) chk[a[i] % 3] = 1;
    long long ans = 0;
    if (!chk[1] && !chk[2]) {
      for (long long i = 1; i <= n; i++) ans = max(a[i] / 3, ans);
    } else if (chk[1] && !chk[2]) {
      for (long long i = 1; i <= n; i++) ans = max(ans, a[i] / 3);
      ans++;
    } else if (!chk[1] && chk[2]) {
      for (long long i = 1; i <= n; i++) ans = max(ans, a[i] / 3);
      ans++;
    } else {
      long long ff = 0;
      for (long long i = 1; i <= n; i++) {
        if (a[i] == 1) ff = 1;
      }
      if (ff) {
        for (long long i = 1; i <= n; i++) {
          if (a[i] % 3 == 0)
            ans = max(ans, a[i] / 3 - 1);
          else
            ans = max(ans, a[i] / 3);
        }
        ans += 2;
      } else {
        long long hubo2 = 0;
        for (long long i = 1; i <= n; i++) {
          if (a[i] % 3 == 0)
            hubo2 = max(hubo2, a[i] / 3 - 1);
          else
            hubo2 = max(hubo2, a[i] / 3);
        }
        hubo2 += 2;
        long long hubo = 0;
        for (long long i = 1; i <= n; i++) {
          if (a[i] % 3 == 1)
            hubo = max(hubo, (a[i] - 4) / 3);
          else
            hubo = max(hubo, a[i] / 3);
        }
        hubo += 2;
        for (long long i = 1; i <= n; i++) {
          if (a[i] % 3 == 1)
            ans = max(ans, (a[i] - 4) / 3);
          else if (a[i] % 3 == 0)
            ans = max(ans, a[i] / 3 - 2);
          else
            ans = max(ans, (a[i] - 2) / 3);
        }
        ans += 3;
        ans = min(ans, hubo);
        ans = min(ans, hubo2);
      }
    }
    printf("%lld\n", ans);
  }
  return 0;
}

```