#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--;
  }
}

### Prompt

In cpp, your task is to solve 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;
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--;
  }
}

```