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.
1
1
6006

2002

#include <bits/stdc++.h>

#define boost ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0)
#define watch(x) cout << (#x) << " : " << x << '\n'
#define pb push_back
#define int long long
#define all(x) (x).begin(), (x).end()
#define rall(x) (x).rbegin(), (x).rend()
#define ll long long
#define point pair<ll, ll>

using namespace std;

const ll INF = 1e17 + 128;

vector <ll> a;
int n;

bool check(ll o, ll b, ll c) {
    for (ll i = 0; i < n; i++) {
        ll cur = a[i];
        ll x = min(cur / 3, c);
        cur -= x*3;
        if (!cur) continue;
        if (cur == 1) {
            if (o)
                continue;
            bool can = false;
            for (int t = 0; t < 3; t++) {
                cur = a[i];
                x = min(((cur-3*t)/3), c-t);
                if (x < 0) continue;
                cur -= 3*x;
                if (cur < 0) continue;
                if (!cur || (cur % 2 == 0 && cur / 2 <= b)) {
                    can = true;
                    break;
                }
            }
            if (can)
                continue;
            return false;
        }
        if (cur == 2) {
            if (b)
                continue;
            bool can = false;
            for (int t = 0; t < 3; t++) {
                cur = a[i];
                x = min(((cur-3*t)/3), c-t);
                if (x < 0) continue;
                cur -= 3*x;
                if (cur < 0) continue;
                if (!cur || cur <= o) {
                    can = true;
                    break;
                }
            }
            if (can)
                continue;
            return false;
        }
        if (cur > 2) {
            bool nice = false;
            for (int i = 0; i <= o; i++)
                for (int j = 0; j <= b; j++)
                    if (i + 2*j == cur)
                        nice = true;
            if (nice) continue;
            return false;
        }
    }
    return true;
}

void test_case() {
    cin >> n;
    a.assign(n, 0);
    for (auto &x : a)
        cin >> x;
    int ans = INF;
    for (int ones = 0; ones <= 3; ones++)
        for (int twos = 0; twos <= 3; twos++) {
            ll l = 0, r = 5e9 + 256;
            while (l + 1 < r) {
                ll mid = (l + r) >> 1;
                if (check(ones, twos, mid))
                    r = mid;
                else
                    l = mid;
            }
            ans = min(ans, ones+twos+(r - !check(ones, twos, r)));
        }
    cout << ans << '\n';
}

main () {
    boost;
    int t = 1;
    cin >> t;
    while (t--)
        test_case();
    return 0;
}
