#include <bits/stdc++.h>
using namespace std;
const int inf = 1e9;
const int maxu = 200005;
void google(int t) { cout << "Case #" << t << ": "; }
int gcd(int a, int b) {
  if (b > a) {
    return gcd(b, a);
  }
  if (b == 0) {
    return a;
  }
  return gcd(b, a % b);
}
int fun(int a, int b) {
  if (b == 0)
    return 1;
  else if (a == 1)
    return 1;
  int temp = fun(a, b / 2) % 1000000007;
  temp = (temp * temp) % 1000000007;
  if (b % 2 == 1) temp = (temp * a) % 1000000007;
  return temp % 1000000007;
}
int inv(int a, int b) { return (a * fun(b, 1000000007 - 2)) % 1000000007; }
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int i, j, k = 0, n, m, l = 0;
  int t = 1;
  int T = 1;
  cin >> T;
  for (t = 1; t <= T; t++) {
    cin >> n;
    int a[n];
    int maxi = 0;
    for (i = 0; i < n; i++) {
      cin >> a[i];
      maxi = max(maxi, a[i]);
    }
    int c3max = (maxi / 3) + 3;
    int ans = inf;
    for (int c1 = 0; c1 <= 4; c1++) {
      for (int c2 = 0; c2 <= 4; c2++) {
        for (int c3 = max(0, c3max - 6); c3 <= c3max; c3++) {
          int cnt = c1 + c2 + c3;
          if (cnt >= ans) continue;
          int can = 1;
          for (i = 0; i < n; i++) {
            int can2 = 0;
            for (j = 0; j <= c1; j++) {
              for (k = 0; k <= c2; k++) {
                int x = a[i] - j - k * 2;
                if (x == 0) {
                  can2 = 1;
                  break;
                }
                if (x < 0) continue;
                if ((x % 3 == 0) && (x <= 3 * c3)) {
                  can2 = 1;
                  break;
                }
              }
              if (can2 == 1) break;
            }
            if (can2 == 0) {
              can = 0;
              break;
            }
          }
          if (can) ans = cnt;
        }
      }
    }
    cout << ans << "\n";
  }
  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 = 1e9;
const int maxu = 200005;
void google(int t) { cout << "Case #" << t << ": "; }
int gcd(int a, int b) {
  if (b > a) {
    return gcd(b, a);
  }
  if (b == 0) {
    return a;
  }
  return gcd(b, a % b);
}
int fun(int a, int b) {
  if (b == 0)
    return 1;
  else if (a == 1)
    return 1;
  int temp = fun(a, b / 2) % 1000000007;
  temp = (temp * temp) % 1000000007;
  if (b % 2 == 1) temp = (temp * a) % 1000000007;
  return temp % 1000000007;
}
int inv(int a, int b) { return (a * fun(b, 1000000007 - 2)) % 1000000007; }
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int i, j, k = 0, n, m, l = 0;
  int t = 1;
  int T = 1;
  cin >> T;
  for (t = 1; t <= T; t++) {
    cin >> n;
    int a[n];
    int maxi = 0;
    for (i = 0; i < n; i++) {
      cin >> a[i];
      maxi = max(maxi, a[i]);
    }
    int c3max = (maxi / 3) + 3;
    int ans = inf;
    for (int c1 = 0; c1 <= 4; c1++) {
      for (int c2 = 0; c2 <= 4; c2++) {
        for (int c3 = max(0, c3max - 6); c3 <= c3max; c3++) {
          int cnt = c1 + c2 + c3;
          if (cnt >= ans) continue;
          int can = 1;
          for (i = 0; i < n; i++) {
            int can2 = 0;
            for (j = 0; j <= c1; j++) {
              for (k = 0; k <= c2; k++) {
                int x = a[i] - j - k * 2;
                if (x == 0) {
                  can2 = 1;
                  break;
                }
                if (x < 0) continue;
                if ((x % 3 == 0) && (x <= 3 * c3)) {
                  can2 = 1;
                  break;
                }
              }
              if (can2 == 1) break;
            }
            if (can2 == 0) {
              can = 0;
              break;
            }
          }
          if (can) ans = cnt;
        }
      }
    }
    cout << ans << "\n";
  }
  return 0;
}

```