#include <bits/stdc++.h>
using namespace std;
const long long sz = 1e5 + 10;
const long long inf = 0x3f3f3f3f;
inline long long read() {
  long long 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(long long *z, long long n) {
  for (long long i = 1; i <= n; i++) printf("%d  ", z[i]);
  puts("");
}
string sa, sb;
long long cnt[sz];
long long n, m;
long long mx;
vector<vector<long long> > p;
long long z[sz];
long long check(long long one, long long two, long long val) {
  long long res = inf;
  for (long long i = 0; i <= one; i++) {
    for (long long j = 0; j <= two; j++) {
      if (val < i + j * 2) continue;
      if ((val - i - j * 2) % 3 == 0) {
        res = min(res, one + two + (val - i - j * 2) / 3);
      }
    }
  }
  return res;
}
long long check(long long one, long long two) {
  long long mx3 = 0;
  for (long long i = 1; i <= n; i++) {
    long long can = check(one, two, z[i]);
    if (can == inf) return inf;
    mx3 = max(mx3, can);
  }
  return mx3;
}
void solve() {
  n = read();
  for (long long i = 1; i <= n; i++) {
    z[i] = read();
  }
  long long mn = inf;
  for (long long one = 0; one <= 2; one++) {
    for (long long two = 0; two <= 2; two++) {
      mn = min(mn, check(one, two));
    }
  }
  cout << mn << endl;
}
signed main() {
  long long t;
  cin >> t;
  while (t--) {
    solve();
  }
}

### 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;
const long long sz = 1e5 + 10;
const long long inf = 0x3f3f3f3f;
inline long long read() {
  long long 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(long long *z, long long n) {
  for (long long i = 1; i <= n; i++) printf("%d  ", z[i]);
  puts("");
}
string sa, sb;
long long cnt[sz];
long long n, m;
long long mx;
vector<vector<long long> > p;
long long z[sz];
long long check(long long one, long long two, long long val) {
  long long res = inf;
  for (long long i = 0; i <= one; i++) {
    for (long long j = 0; j <= two; j++) {
      if (val < i + j * 2) continue;
      if ((val - i - j * 2) % 3 == 0) {
        res = min(res, one + two + (val - i - j * 2) / 3);
      }
    }
  }
  return res;
}
long long check(long long one, long long two) {
  long long mx3 = 0;
  for (long long i = 1; i <= n; i++) {
    long long can = check(one, two, z[i]);
    if (can == inf) return inf;
    mx3 = max(mx3, can);
  }
  return mx3;
}
void solve() {
  n = read();
  for (long long i = 1; i <= n; i++) {
    z[i] = read();
  }
  long long mn = inf;
  for (long long one = 0; one <= 2; one++) {
    for (long long two = 0; two <= 2; two++) {
      mn = min(mn, check(one, two));
    }
  }
  cout << mn << endl;
}
signed main() {
  long long t;
  cin >> t;
  while (t--) {
    solve();
  }
}

```