#include <bits/stdc++.h>
using namespace std;
char ch[1010000];
long long x[1010000];
long long g[10];
vector<long long> vt, ans;
signed main() {
  long long T, n, m, cnt, num, k;
  scanf("%lld", &T);
  while (T--) {
    scanf("%lld", &n);
    long long mmax = 0;
    for (long long i = 1; i <= n; i++) {
      scanf("%lld", &x[i]);
      g[x[i] % 3]++;
      mmax = max(x[i], mmax);
    }
    if (mmax == 1) {
      printf("1\n");
      continue;
    } else if (mmax == 2) {
      long long flag = 0;
      for (long long i = 1; i <= n; i++) {
        if (x[i] == 1) {
          flag = 1;
        }
      }
      if (flag) {
        printf("2\n");
      } else {
        printf("1\n");
      }
      continue;
    } else if (mmax == 3) {
      long long flag = 0;
      for (long long i = 1; i <= n; i++) {
        if (x[i] != 3) {
          flag = 1;
        }
      }
      if (flag) {
        printf("2\n");
      } else {
        printf("1\n");
      }
      continue;
    }
    long long t = mmax % 3;
    long long cnt = (mmax + 2) / 3;
    if (t == 0) {
      long long flag = 0;
      for (long long i = 1; i <= n; i++) {
        if (x[i] % 3 != 0) {
          flag = 1;
          break;
        }
      }
      if (flag) {
        cnt++;
      }
    } else if (t == 1) {
      long long flag = 0;
      for (long long i = 1; i <= n; i++) {
        if (x[i] % 3 == 2) {
          flag = 1;
          break;
        }
      }
      if (flag) {
        for (long long i = 1; i <= n; i++) {
          if (x[i] == 1 || x[i] == mmax - 1) {
            flag = 0;
            break;
          }
        }
        if (flag == 0) {
          cnt++;
        }
      }
    } else if (t == 2) {
      long long flag = 0;
      for (long long i = 1; i <= n; i++) {
        if (x[i] % 3 == 1) {
          flag = 1;
          break;
        }
      }
      if (flag) {
        cnt++;
      }
    }
    printf("%lld\n", cnt);
  }
  return 0;
}

### Prompt

Develop a solution in Cpp to the problem described below:

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;
char ch[1010000];
long long x[1010000];
long long g[10];
vector<long long> vt, ans;
signed main() {
  long long T, n, m, cnt, num, k;
  scanf("%lld", &T);
  while (T--) {
    scanf("%lld", &n);
    long long mmax = 0;
    for (long long i = 1; i <= n; i++) {
      scanf("%lld", &x[i]);
      g[x[i] % 3]++;
      mmax = max(x[i], mmax);
    }
    if (mmax == 1) {
      printf("1\n");
      continue;
    } else if (mmax == 2) {
      long long flag = 0;
      for (long long i = 1; i <= n; i++) {
        if (x[i] == 1) {
          flag = 1;
        }
      }
      if (flag) {
        printf("2\n");
      } else {
        printf("1\n");
      }
      continue;
    } else if (mmax == 3) {
      long long flag = 0;
      for (long long i = 1; i <= n; i++) {
        if (x[i] != 3) {
          flag = 1;
        }
      }
      if (flag) {
        printf("2\n");
      } else {
        printf("1\n");
      }
      continue;
    }
    long long t = mmax % 3;
    long long cnt = (mmax + 2) / 3;
    if (t == 0) {
      long long flag = 0;
      for (long long i = 1; i <= n; i++) {
        if (x[i] % 3 != 0) {
          flag = 1;
          break;
        }
      }
      if (flag) {
        cnt++;
      }
    } else if (t == 1) {
      long long flag = 0;
      for (long long i = 1; i <= n; i++) {
        if (x[i] % 3 == 2) {
          flag = 1;
          break;
        }
      }
      if (flag) {
        for (long long i = 1; i <= n; i++) {
          if (x[i] == 1 || x[i] == mmax - 1) {
            flag = 0;
            break;
          }
        }
        if (flag == 0) {
          cnt++;
        }
      }
    } else if (t == 2) {
      long long flag = 0;
      for (long long i = 1; i <= n; i++) {
        if (x[i] % 3 == 1) {
          flag = 1;
          break;
        }
      }
      if (flag) {
        cnt++;
      }
    }
    printf("%lld\n", cnt);
  }
  return 0;
}

```