#include <bits/stdc++.h>
using namespace std;
inline long long read() {
  long long n = 0, f = 1, ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    n = n * 10 + ch - '0';
    ch = getchar();
  }
  return n * f;
}
long long a[200005];
bool kd[8];
signed main() {
  long long t, n, sth, ans;
  bool flag;
  t = read();
  for (long long greg = 1; greg <= t; greg++) {
    n = read();
    for (long long i = 1; i <= n; i++) a[i] = read();
    ans = 1000000000;
    for (long long i = 0; i <= 1; i++) {
      for (long long j = 0; j <= 2; j++) {
        for (long long k = 0; k <= 5; k++) kd[k] = false;
        kd[0] = kd[i] = true;
        if (j >= 1) kd[i + 2] = kd[2] = true;
        if (j >= 2) kd[i + 4] = kd[4] = true;
        sth = 0;
        flag = true;
        for (long long ti = 1; ti <= n; ti++) {
          for (long long tj = 5; tj >= 0; tj--) {
            if (a[ti] >= tj && kd[tj] == true && a[ti] % 3 == tj % 3) {
              sth = max(sth, (a[ti] - tj) / 3);
              break;
            }
            if (tj == 0) flag = false;
          }
          if (flag == false) break;
        }
        if (flag == false) continue;
        ans = min(ans, sth + i + j);
      }
    }
    printf("%lld\n", ans);
  }
  return 0;
}

### Prompt

Construct a CPP code solution to the problem outlined:

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;
inline long long read() {
  long long n = 0, f = 1, ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') {
    n = n * 10 + ch - '0';
    ch = getchar();
  }
  return n * f;
}
long long a[200005];
bool kd[8];
signed main() {
  long long t, n, sth, ans;
  bool flag;
  t = read();
  for (long long greg = 1; greg <= t; greg++) {
    n = read();
    for (long long i = 1; i <= n; i++) a[i] = read();
    ans = 1000000000;
    for (long long i = 0; i <= 1; i++) {
      for (long long j = 0; j <= 2; j++) {
        for (long long k = 0; k <= 5; k++) kd[k] = false;
        kd[0] = kd[i] = true;
        if (j >= 1) kd[i + 2] = kd[2] = true;
        if (j >= 2) kd[i + 4] = kd[4] = true;
        sth = 0;
        flag = true;
        for (long long ti = 1; ti <= n; ti++) {
          for (long long tj = 5; tj >= 0; tj--) {
            if (a[ti] >= tj && kd[tj] == true && a[ti] % 3 == tj % 3) {
              sth = max(sth, (a[ti] - tj) / 3);
              break;
            }
            if (tj == 0) flag = false;
          }
          if (flag == false) break;
        }
        if (flag == false) continue;
        ans = min(ans, sth + i + j);
      }
    }
    printf("%lld\n", ans);
  }
  return 0;
}

```