#include <bits/stdc++.h>
using namespace std;
mt19937 gene(233);
inline char GET_CHAR() {
  const int maxn = 131072;
  static char buf[maxn], *p1 = buf, *p2 = buf;
  return p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, maxn, stdin), p1 == p2)
             ? EOF
             : *p1++;
}
inline int getInt() {
  int res(0);
  char c = getchar();
  while (c < '0') c = getchar();
  while (c >= '0') {
    res = res * 10 + (c - '0');
    c = getchar();
  }
  return res;
}
inline long long fastpo(long long x, long long n, long long mod) {
  long long res(1);
  while (n) {
    if (n & 1) {
      res = res * (long long)x % mod;
    }
    x = x * (long long)x % mod;
    n /= 2;
  }
  return res;
}
long long gcd(long long a, long long b) { return b ? gcd(b, a % b) : a; }
inline string itoa(int x, int width = 0) {
  string res;
  if (x == 0) res.push_back('0');
  while (x) {
    res.push_back('0' + x % 10);
    x /= 10;
  }
  while ((int)res.size() < width) res.push_back('0');
  reverse(res.begin(), res.end());
  return res;
}
const int _B = 131072;
char buf[_B];
int _bl = 0;
inline void flush() {
  fwrite(buf, 1, _bl, stdout);
  _bl = 0;
}
__inline void _putchar(char c) {
  if (_bl == _B) flush();
  buf[_bl++] = c;
}
inline void print(long long x, char c) {
  static char tmp[20];
  int l = 0;
  if (!x)
    tmp[l++] = '0';
  else {
    while (x) {
      tmp[l++] = x % 10 + '0';
      x /= 10;
    }
  }
  for (int i = l - 1; i >= 0; i--) _putchar(tmp[i]);
  _putchar(c);
}
struct P {
  double x, y;
};
const int N = 300033;
const int LOG = 20;
const int mod = 1e9 + 7;
const int inf = 1e9 + 7;
int n, m;
int dx[4] = {1, 0, -1, 0};
int dy[4] = {0, 1, 0, -1};
int a[N], f[N];
int main() {
  int t;
  scanf("%d", &t);
  for (int qq = 1; qq <= t; qq++) {
    int n;
    scanf("%d", &n);
    for (int i = 1; i <= n; i++) {
      scanf("%d", &a[i]);
    }
    int ans = inf;
    for (int o = 0; o <= 6; o++) {
      for (int t = 0; t <= 6; t++) {
        fill(f, f + o + t * 2 + 1, false);
        for (int i = 0; i <= o; i++) {
          for (int j = 0; j <= t; j++) {
            f[i + j * 2] = true;
          }
        }
        int mx = 0;
        for (int i = 1; i <= n; i++) {
          int tmp = inf;
          for (int j = o + t * 2; j >= 0; j--) {
            if (f[j] && a[i] >= j && (a[i] - j) % 3 == 0) {
              tmp = (a[i] - j) / 3;
              break;
            }
          }
          mx = max(mx, tmp);
        }
        ans = min(ans, o + t + mx);
      }
    }
    printf("%d\n", ans);
  }
}

### 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;
mt19937 gene(233);
inline char GET_CHAR() {
  const int maxn = 131072;
  static char buf[maxn], *p1 = buf, *p2 = buf;
  return p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, maxn, stdin), p1 == p2)
             ? EOF
             : *p1++;
}
inline int getInt() {
  int res(0);
  char c = getchar();
  while (c < '0') c = getchar();
  while (c >= '0') {
    res = res * 10 + (c - '0');
    c = getchar();
  }
  return res;
}
inline long long fastpo(long long x, long long n, long long mod) {
  long long res(1);
  while (n) {
    if (n & 1) {
      res = res * (long long)x % mod;
    }
    x = x * (long long)x % mod;
    n /= 2;
  }
  return res;
}
long long gcd(long long a, long long b) { return b ? gcd(b, a % b) : a; }
inline string itoa(int x, int width = 0) {
  string res;
  if (x == 0) res.push_back('0');
  while (x) {
    res.push_back('0' + x % 10);
    x /= 10;
  }
  while ((int)res.size() < width) res.push_back('0');
  reverse(res.begin(), res.end());
  return res;
}
const int _B = 131072;
char buf[_B];
int _bl = 0;
inline void flush() {
  fwrite(buf, 1, _bl, stdout);
  _bl = 0;
}
__inline void _putchar(char c) {
  if (_bl == _B) flush();
  buf[_bl++] = c;
}
inline void print(long long x, char c) {
  static char tmp[20];
  int l = 0;
  if (!x)
    tmp[l++] = '0';
  else {
    while (x) {
      tmp[l++] = x % 10 + '0';
      x /= 10;
    }
  }
  for (int i = l - 1; i >= 0; i--) _putchar(tmp[i]);
  _putchar(c);
}
struct P {
  double x, y;
};
const int N = 300033;
const int LOG = 20;
const int mod = 1e9 + 7;
const int inf = 1e9 + 7;
int n, m;
int dx[4] = {1, 0, -1, 0};
int dy[4] = {0, 1, 0, -1};
int a[N], f[N];
int main() {
  int t;
  scanf("%d", &t);
  for (int qq = 1; qq <= t; qq++) {
    int n;
    scanf("%d", &n);
    for (int i = 1; i <= n; i++) {
      scanf("%d", &a[i]);
    }
    int ans = inf;
    for (int o = 0; o <= 6; o++) {
      for (int t = 0; t <= 6; t++) {
        fill(f, f + o + t * 2 + 1, false);
        for (int i = 0; i <= o; i++) {
          for (int j = 0; j <= t; j++) {
            f[i + j * 2] = true;
          }
        }
        int mx = 0;
        for (int i = 1; i <= n; i++) {
          int tmp = inf;
          for (int j = o + t * 2; j >= 0; j--) {
            if (f[j] && a[i] >= j && (a[i] - j) % 3 == 0) {
              tmp = (a[i] - j) / 3;
              break;
            }
          }
          mx = max(mx, tmp);
        }
        ans = min(ans, o + t + mx);
      }
    }
    printf("%d\n", ans);
  }
}

```