description: 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.
solution: #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);
    }
  }
  