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;
const int sz = 5e5 + 10;
const int inf = 0x3f3f3f3f;
inline int read() {
  int f = 1, x = 0;
  char ch = getchar();
  for (; !isdigit(ch); ch = getchar())
    if (ch == '-') f = -1;
  for (; isdigit(ch); ch = getchar()) x = x * 10 + ch - 48;
  return f * x;
}
void print(int *z, int n) {
  for (int i = 1; i <= n; i++) printf("%d  ", z[i]);
  puts("");
}
int z[sz];
int solve() {
  int n = read();
  bool one = 0, two = 0, trone = 0;
  int mx = 0;
  for (int i = 1; i <= n; i++) {
    z[i] = read();
    if (z[i] == 1) trone = 1;
    if (z[i] % 3 == 1)
      one = 1;
    else if (z[i] % 3 == 2)
      two = 1;
    mx = max(mx, z[i]);
  }
  if (mx % 3 == 0) {
    if (one || two) return mx / 3 + 1;
    return mx / 3;
  } else if (mx % 3 == 2) {
    if (one)
      return mx / 3 + 2;
    else
      return mx / 3 + 1;
  } else {
    if (!two)
      return mx / 3 + 1;
    else {
      if (!trone) return mx / 3 + 1;
      return mx / 3 + 2;
    }
  }
}
signed main() {
  int t;
  cin >> t;
  while (t--) {
    cout << solve() << endl;
  }
}
