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.
1
1
6006

2002

#include <bits/stdc++.h>
using namespace std;
int len(long long int x) {
  int i = 0;
  while (x >= 10) {
    i++;
    x /= 10;
  }
  return i;
}
int main() {
  int t;
  cin >> t;
  while (t--) {
    vector<long long int> v1(4);
    vector<long long int> v2(4);
    for (int i = 0; i < 4; i++) {
      v1[i] = 0;
      v2[i] = 0;
    }
    int n;
    cin >> n;
    vector<long long int> v(n);
    for (int i = 0; i < n; i++) {
      int x;
      cin >> x;
      v[i] = x;
    }
    sort(v.begin(), v.end());
    for (int i = 0; i < n; i++) {
      if (i == n - 1) {
        if (v[i] >= 3) {
          if (v[i] % 3 == 0) {
            v1[3] = max(v1[3], v[i] / 3);
            v2[3] = max(v2[3], v[i] / 3);
            if (v1[1] && v1[2]) {
              cout << "Tafi" << endl;
              v1[3]--;
            }
            if (v2[1] && v2[2]) {
              v2[3]--;
            }
          } else if (v[i] % 3 == 1) {
            v2[2] = max(v2[2], 2LL);
            v2[3] = max(v2[3], v[i] / 3 - 1);
            v1[1] = max(v1[1], 1LL);
            v1[3] = max(v1[3], v[i] / 3);
          } else if (v[i] % 3 == 2) {
            v1[3] = max(v1[3], v[i] / 3);
            v1[2] = max(v1[2], 1LL);
            if (v1[2] > 1 && v1[1]) {
              v1[3]--;
            }
            v2[3] = max(v2[3], v[i] / 3);
            v2[2] = max(v2[2], 1LL);
            if (v2[2] > 1 && v2[1]) {
              v2[3]--;
            }
          }
        }
      } else if (v[i] >= 3) {
        if (v[i] % 3 == 1) {
          v2[2] = max(v2[2], 2LL);
          v2[3] = max(v2[3], v[i] / 3 - 1);
          v1[1] = max(v1[1], 1LL);
          v1[3] = max(v1[3], v[i] / 3);
        } else if (v[i] % 3 == 2) {
          v1[3] = max(v1[3], v[i] / 3);
          v1[2] = max(v1[2], 1LL);
          v2[3] = max(v2[3], v[i] / 3);
          v2[2] = max(v2[2], 1LL);
        } else {
          v1[3] = max(v1[3], v[i] / 3);
          v2[3] = max(v2[3], v[i] / 3);
        }
      } else if (v[i] >= 2) {
        if (v1[1] < 2) {
          v1[2] = max(v1[2], 1LL);
        }
        if (v2[1] < 2) {
          v2[2] = max(v2[2], 1LL);
        }
      } else {
        v1[1] = max(v1[1], 1LL);
        v2[1] = max(v2[1], 1LL);
      }
    }
    long long int m = v[n - 1];
    long long int sum1 = 0, sum2 = 0;
    for (int i = 0; i < 4; i++) {
      sum1 += v1[i];
      sum2 += v2[i];
    }
    if (sum1 < sum2) {
      cout << sum1 << endl;
    } else {
      cout << sum2 << endl;
    }
  }
}
