#include <bits/stdc++.h>
using namespace std;
inline int read() {
  int s = 0, w = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') w = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') s = s * 10 + ch - '0', ch = getchar();
  return s * w;
}
int n, a[1005];
int f(int x, int y) {
  int s = 0;
  for (int i = (1); i <= (n); i++) {
    int p = 2100000000;
    for (int j = (0); j <= (x); j++)
      for (int k = (0); k <= (y); k++) {
        int w = j + k * 2;
        if (a[i] >= w && (a[i] - w) % 3 == 0) p = min(p, (a[i] - w) / 3);
      }
    s = max(s, p);
  }
  s += x + y;
  return s;
}
int main() {
  int T = read();
  while (T--) {
    n = read();
    for (int i = (1); i <= (n); i++) a[i] = read();
    int ans = 2100000000;
    for (int i = (0); i <= (2); i++)
      for (int j = (0); j <= (2); j++) ans = min(ans, f(i, j));
    printf("%d\n", ans);
  }
  return 0;
}

### 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;
inline int read() {
  int s = 0, w = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') w = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') s = s * 10 + ch - '0', ch = getchar();
  return s * w;
}
int n, a[1005];
int f(int x, int y) {
  int s = 0;
  for (int i = (1); i <= (n); i++) {
    int p = 2100000000;
    for (int j = (0); j <= (x); j++)
      for (int k = (0); k <= (y); k++) {
        int w = j + k * 2;
        if (a[i] >= w && (a[i] - w) % 3 == 0) p = min(p, (a[i] - w) / 3);
      }
    s = max(s, p);
  }
  s += x + y;
  return s;
}
int main() {
  int T = read();
  while (T--) {
    n = read();
    for (int i = (1); i <= (n); i++) a[i] = read();
    int ans = 2100000000;
    for (int i = (0); i <= (2); i++)
      for (int j = (0); j <= (2); j++) ans = min(ans, f(i, j));
    printf("%d\n", ans);
  }
  return 0;
}

```