#include<bits/stdc++.h>
using namespace std;
int64_t a[200005], n, m;
int64_t cnt[5][200005];
array<int64_t, 100005> h, h2;
array<int64_t, 5> b, b2;
void solve(){
    cin >> n;
    for(int i = 1; i <= n; i++) cin >> a[i];
    for(int i = n + 1; i <= 2 * n; i++) a[i] = a[i - n];
    for(int i = 2; i <= 2 * n; i++) a[i] += a[i - 1];
    for(int i = 1; i <= 2 * n; i++){
        cnt[1][i] = a[i] - a[i - 1] == 1;
        cnt[2][i] = a[i] - a[i - 1] == 2;
        cnt[3][i] = a[i] - a[i - 1] == 3;
        cnt[1][i] += cnt[1][i - 1];
        cnt[2][i] += cnt[2][i - 1];
        cnt[3][i] += cnt[3][i - 1];
    }
    cin >> m;
    int64_t s = 0;
    for(int i = 1; i <= m; i++) cin >> h[i], s += h[i];
    int64_t L = 1, R = s + 1;
    while(L < R){
        int64_t mid = (L + R) / 2;
        int64_t dmg = mid / n * a[n] + a[mid % n];
        // printf("L = %lld, R = %lld, mid = %lld, dmg = %lld, s = %lld\n", L, R, mid, dmg, s);
        if(dmg >= s) R = mid;
        else L = mid + 1;
    }
    int64_t ans = L, cur = L % n;
    // printf("ans = %lld\n", ans);
    b = {};
    b[1] = L / n * cnt[1][n] + cnt[1][L % n];
    b[2] = L / n * cnt[2][n] + cnt[2][L % n];
    b[3] = L / n * cnt[3][n] + cnt[3][L % n];
    
    sort(&h[1], &h[m + 1]);
    h2 = h, b2 = b;
    while(1){
        // cout << b[1] << ' ' << b[2] << ' ' << b[3] << endl;
        bool ok = 1;
        for(int i = 1; i <= m; i++){
            int need3 = h[i] / 3;
            if(need3 <= b[3]) b[3] -= need3;
            else need3 = b[3];
            h[i] -= need3 * 3;
            
            int need2 = h[i] / 2;
            if(need2 <= b[2]) b[2] -= need2;
            else need2 = b[2];
            h[i] -= need2 * 2;
    
            int need1 = h[i];
            if(need1 <= b[1]) b[1] -= need1;
            else need1 = b[1];
            h[i] -= need1;
            
            if(h[i] > 0){
                need2 = (h[i] + 1) / 2;
                if(need2 <= b[2]) b[2] -= need2;
                else need2 = b[2];
                h[i] -= need2 * 2;
            }
            if(h[i] > 0){
                need3 = (h[i] + 2) / 3;
                if(need3 <= b[3]) b[3] -= need3;
                else need3 = b[3];
                h[i] -= need3 * 3;
            }
            if(h[i] > 0){ ok = 0; break; }
        }
        if(ok) break;
        h = h2;
        ans++;
        b2[a[cur + 1] - a[cur]]++;
        b = b2;
    }
    cout << ans;
    cout << '\n';
}
int main(){
    cin.tie(0);
    int T;
    cin >> T;
    while(T--) solve();
    return 0;
}
/*
要想次数最少，攻击时伤害值尽可能刚好等于怪物的血量。
依次释放技能攻击第1、2、3、……个怪物，
攻击第i（i>1）个怪物进行某次攻击时，
若前面的怪物有伤害溢出，且本次伤害值为1或2，
就可以考虑交换一次伤害以减少伤害溢出
*/