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;
  long long mod = pow(10, 9) + 7;
  int gcdExtended(int a, int b, int *x, int *y);
  int modInverse(int b, int m) {
    int x, y;
    int g = gcdExtended(b, m, &x, &y);
    if (g != 1) return -1;
    return (x % m + m) % m;
  }
  int gcdExtended(int a, int b, int *x, int *y) {
    if (a == 0) {
      *x = 0, *y = 1;
      return b;
    }
    int x1, y1;
    int gcd = gcdExtended(b % a, a, &x1, &y1);
    *x = y1 - (b / a) * x1;
    *y = x1;
    return gcd;
  }
  long long sumbincoef(int N, int k) {
    long long bincoef = 1, sum = 1;
    int i;
    for (i = 1; i < k; i++) {
      long long val = ((N - i + 1) % mod * modInverse(i, mod)) % mod;
      bincoef = (bincoef % mod * val) % mod;
      if (i % 2 == 0) sum = (sum + bincoef) % mod;
    }
    return sum % mod;
  }
  const int N = 300000 + 10;
  long long factorialNumInverse[N + 1];
  long long naturalNumInverse[N + 1];
  long long fact[N + 1];
  void InverseofNumber(long long p) {
    naturalNumInverse[0] = naturalNumInverse[1] = 1;
    for (int i = 2; i <= N; i++)
      naturalNumInverse[i] = naturalNumInverse[p % i] * (p - p / i) % p;
  }
  void InverseofFactorial(long long p) {
    factorialNumInverse[0] = factorialNumInverse[1] = 1;
    for (int i = 2; i <= N; i++)
      factorialNumInverse[i] =
          (naturalNumInverse[i] * factorialNumInverse[i - 1]) % p;
  }
  void factorial(long long p) {
    fact[0] = 1;
    for (int i = 1; i <= N; i++) {
      fact[i] = (fact[i - 1] * i) % p;
    }
  }
  long long Binomial(long long N, long long R, long long p) {
    long long ans =
        ((fact[N] * factorialNumInverse[R]) % p * factorialNumInverse[N - R]) % p;
    return ans;
  }
  vector<int> parent(400000 + 1), ran(400000 + 1);
  void make_set(int v) {
    parent[v] = v;
    ran[v] = 0;
  }
  int find_set(int a) {
    if (parent[a] == a) {
      return a;
    }
    return parent[a] = find_set(parent[a]);
  }
  int union_sets(int a, int b) {
    a = find_set(a);
    b = find_set(b);
    if (ran[a] < ran[b]) {
      parent[a] = b;
    } else if (ran[a] > ran[b]) {
      parent[b] = a;
    } else {
      parent[b] = a;
      ran[a]++;
    }
  }
  int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);
    mt19937_64 rng(std::chrono::steady_clock::now().time_since_epoch().count());
    long long p = mod;
    InverseofNumber(p);
    InverseofFactorial(p);
    factorial(p);
    int t = 1;
    cin >> t;
    while (t > 0) {
      int n;
      cin >> n;
      vector<long long> a(n), b(n);
      for (int i = 0; i < n; i++) cin >> a[i];
      long long ans = 1e9;
      for (long long c1 = 0; c1 <= 3; c1++) {
        for (long long c2 = 0; c2 <= 3; c2++) {
          long long lo = 0;
          long long hi = 1e9;
          while (lo <= hi) {
            long long mid = (lo + hi) / 2;
            long long c3 = mid;
            bool ok = true;
            for (int i = 0; i < n; i++) {
              bool lok = false;
              for (long long cnt1 = 0; cnt1 <= c1; cnt1++) {
                for (long long cnt2 = 0; cnt2 <= c2; cnt2++) {
                  long long now = a[i] - cnt1 - 2 * cnt2;
                  if (now % 3 == 0 && now >= 0) {
                    if (now / 3 <= c3) {
                      lok = true;
                      break;
                    }
                  }
                }
                if (lok) {
                  break;
                }
              }
              if (!lok) {
                ok = false;
                break;
              }
            }
            if (ok) {
              ans = min(ans, c1 + c2 + mid);
              hi = mid - 1;
            } else {
              lo = mid + 1;
            }
          }
        }
      }
      cout << ans << "\n";
      t--;
    }
  }
  