#include <bits/stdc++.h>
int int_cmp(const void *a, const void *b) { return *((int *)a) - *((int *)b); }
int unique(int a[], int size) {
  int i, k;
  if (size == 0) return 0;
  for (i = 1; i < size; i++)
    if (a[i] == a[i - 1]) break;
  for (k = i, i++; i < size; i++)
    if (a[i] != a[i - 1]) a[k++] = a[i];
  return k;
}
int main(int argc, char *argv[]) {
  int i, T, n, onec, twoc, q, r, ans;
  int cost[128];
  scanf("%d", &T);
  while (T-- > 0) {
    scanf("%d", &n);
    onec = twoc = 0;
    for (i = 0; i < n; i++) {
      scanf("%d", &cost[i]);
      onec |= (cost[i] % 3) == 1;
      twoc |= (cost[i] % 3) == 2;
    }
    qsort(cost, n, sizeof(cost[0]), int_cmp);
    n = unique(cost, n);
    q = cost[n - 1] / 3;
    if (onec == 0 && twoc == 0)
      ans = q;
    else if (onec == 0 || twoc == 0)
      ans = q + 1;
    else {
      r = cost[n - 1] % 3;
      if (r == 0)
        ans = q + 1;
      else if (r == 2)
        ans = q + 2;
      else if (cost[0] == 1 || cost[n - 2] == cost[n - 1] - 1)
        ans = q + 2;
      else
        ans = q + 1;
    }
    printf("%d\n", ans);
  }
  return 0;
}

### Prompt

Please provide a CPP coded solution 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>
int int_cmp(const void *a, const void *b) { return *((int *)a) - *((int *)b); }
int unique(int a[], int size) {
  int i, k;
  if (size == 0) return 0;
  for (i = 1; i < size; i++)
    if (a[i] == a[i - 1]) break;
  for (k = i, i++; i < size; i++)
    if (a[i] != a[i - 1]) a[k++] = a[i];
  return k;
}
int main(int argc, char *argv[]) {
  int i, T, n, onec, twoc, q, r, ans;
  int cost[128];
  scanf("%d", &T);
  while (T-- > 0) {
    scanf("%d", &n);
    onec = twoc = 0;
    for (i = 0; i < n; i++) {
      scanf("%d", &cost[i]);
      onec |= (cost[i] % 3) == 1;
      twoc |= (cost[i] % 3) == 2;
    }
    qsort(cost, n, sizeof(cost[0]), int_cmp);
    n = unique(cost, n);
    q = cost[n - 1] / 3;
    if (onec == 0 && twoc == 0)
      ans = q;
    else if (onec == 0 || twoc == 0)
      ans = q + 1;
    else {
      r = cost[n - 1] % 3;
      if (r == 0)
        ans = q + 1;
      else if (r == 2)
        ans = q + 2;
      else if (cost[0] == 1 || cost[n - 2] == cost[n - 1] - 1)
        ans = q + 2;
      else
        ans = q + 1;
    }
    printf("%d\n", ans);
  }
  return 0;
}

```