#include <bits/stdc++.h>
using namespace std;
long long M1 = pow(10, 9) + 7, M2 = 998244353, NX = 9223372036854775807,
          LN = 1e18;
long long binpow(long long x, long long n) {
  long long res = 1;
  while (n) {
    if (n % 2) res = (res * x);
    n /= 2;
    x = (x * x);
  }
  return res;
}
long long modpow(long long x, long long n, long long M) {
  x = x % M;
  long long res = 1;
  while (n) {
    if (n % 2) res = (res * x) % M;
    n /= 2;
    x = (x * x) % M;
  }
  return res;
}
long long gcd(long long a, long long b) {
  if (!b) return a;
  return gcd(b, a % b);
}
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long tc = 1;
  cin >> tc;
  while (tc--) {
    long long n;
    cin >> n;
    vector<long long> v(n);
    long long i, m = 0, cnt[3] = {};
    for (i = 0; i < n; i++) {
      cin >> v[i];
      m = max(m, v[i]);
      cnt[v[i] % 3] = 1;
    }
    long long f = 0, g = 0;
    for (i = 0; i < n; i++) {
      if (v[i] == 1) f = 1;
      if (v[i] == m - 1) g = 1;
    }
    if (m % 3 == 0)
      cout << m / 3 + (cnt[1] | cnt[2]) << "\n";
    else if (m % 3 == 1) {
      if (!cnt[2] || (cnt[2] && !f && !g))
        cout << m / 3 + 1 << "\n";
      else
        cout << m / 3 + 2 << "\n";
    } else
      cout << m / 3 + 1 + cnt[1] << "\n";
  }
  return 0;
}

### Prompt

Create a solution in CPP for 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 M1 = pow(10, 9) + 7, M2 = 998244353, NX = 9223372036854775807,
          LN = 1e18;
long long binpow(long long x, long long n) {
  long long res = 1;
  while (n) {
    if (n % 2) res = (res * x);
    n /= 2;
    x = (x * x);
  }
  return res;
}
long long modpow(long long x, long long n, long long M) {
  x = x % M;
  long long res = 1;
  while (n) {
    if (n % 2) res = (res * x) % M;
    n /= 2;
    x = (x * x) % M;
  }
  return res;
}
long long gcd(long long a, long long b) {
  if (!b) return a;
  return gcd(b, a % b);
}
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long tc = 1;
  cin >> tc;
  while (tc--) {
    long long n;
    cin >> n;
    vector<long long> v(n);
    long long i, m = 0, cnt[3] = {};
    for (i = 0; i < n; i++) {
      cin >> v[i];
      m = max(m, v[i]);
      cnt[v[i] % 3] = 1;
    }
    long long f = 0, g = 0;
    for (i = 0; i < n; i++) {
      if (v[i] == 1) f = 1;
      if (v[i] == m - 1) g = 1;
    }
    if (m % 3 == 0)
      cout << m / 3 + (cnt[1] | cnt[2]) << "\n";
    else if (m % 3 == 1) {
      if (!cnt[2] || (cnt[2] && !f && !g))
        cout << m / 3 + 1 << "\n";
      else
        cout << m / 3 + 2 << "\n";
    } else
      cout << m / 3 + 1 + cnt[1] << "\n";
  }
  return 0;
}

```