Description

The 2018 World Cup will be hosted in Russia. 32 national teams will be divided into 8 groups. Each group consists of 4 teams. In group matches, each pair (unordered) of teams in the group will have a match. Top 2 teams with the highest score in each group will advance to eighth-finals. Winners of each eighth-final will advance to quarter-finals. Then, the winners of each quarter-final will advance to semi-finals. Eventually, the World Champion will be the winner of the World Final which is played between the two winners of the semi-finals.

Each match is labeled with a match ID sequenced from 1 to 63, with group matches followed by eighth-final matches followed by quarter-final matches followed by semi-finals matches and finally the final match.

Zhuojie is going to watch the 2018 World Cup. Since the World Champion of ACM-ICPC is very rich, he decides to spend 0.01% of his daily salary to buy tickets. However, there are only match IDs on the tickets and the prices are missing. Can you calculate how much Google pays Zhuojie every workday? Note that Zhuojie can buy multiple tickets for one match.

Input

The input starts with one line containing exactly one integer T, the number of test cases.

Each test case contains 3 lines. The first line contains 5 integers, indicating the ticket price for group match, eighth-final match, quarter-final match, semi-final match and the final match. The second line contains one integer N, the number of tickets Zhuojie buys. The third line contains N integers, each indicating the match ID on the ticket.

• 1 ≤ T ≤ 100.
• 1 ≤ N ≤ 105.
• .
• .

Output

For each test case, output one line containing "Case #x: y" where x is the test case number (starting from 1) and y is daily salary of Zhuojie.

Sample Input
Input 
1
11 12 13 14 15
2
1 49
Output 
Case #1: 230000

题意：有五种比赛（编号1-63），预选赛（1-48），16进8（49-56），8进4（57-60），4进2（61，62），2进1（63），给出着五种比赛的票价，问这个人买n张票（给出要买票的编号），他花了工资的0.01%，问这个人工资是多少。

/***
 * ┌───┐   ┌───┬───┬───┬───┐ ┌───┬───┬───┬───┐ ┌───┬───┬───┬───┐ ┌───┬───┬───┐
 * │Esc│   │ F1│ F2│ F3│ F4│ │ F5│ F6│ F7│ F8│ │ F9│F10│F11│F12│ │P/S│S L│P/B│  ┌┐    ┌┐    ┌┐
 * └───┘   └───┴───┴───┴───┘ └───┴───┴───┴───┘ └───┴───┴───┴───┘ └───┴───┴───┘  └┘    └┘    └┘
 * ┌───┬───┬───┬───┬───┬───┬───┬───┬───┬───┬───┬───┬───┬───────┐ ┌───┬───┬───┐ ┌───┬───┬───┬───┐
 * │~ `│! 1│@ 2│# 3│$ 4│% 5│^ 6│& 7│* 8│( 9│) 0│_ -│+ =│ BacSp │ │Ins│Hom│PUp│ │N L│ / │ * │ - │
 * ├───┴─┬─┴─┬─┴─┬─┴─┬─┴─┬─┴─┬─┴─┬─┴─┬─┴─┬─┴─┬─┴─┬─┴─┬─┴─┬─────┤ ├───┼───┼───┤ ├───┼───┼───┼───┤
 * │ Tab │ Q │ W │ E │ R │ T │ Y │ U │ I │ O │ P │{ [│} ]│ | \ │ │Del│End│PDn│ │ 7 │ 8 │ 9 │   │
 * ├─────┴┬──┴┬──┴┬──┴┬──┴┬──┴┬──┴┬──┴┬──┴┬──┴┬──┴┬──┴┬──┴─────┤ └───┴───┴───┘ ├───┼───┼───┤ + │
 * │ Caps │ A │ S │ D │ F │ G │ H │ J │ K │ L │: ;│" '│ Enter  │               │ 4 │ 5 │ 6 │   │
 * ├──────┴─┬─┴─┬─┴─┬─┴─┬─┴─┬─┴─┬─┴─┬─┴─┬─┴─┬─┴─┬─┴─┬─┴────────┤     ┌───┐     ├───┼───┼───┼───┤
 * │ Shift  │ Z │ X │ C │ V │ B │ N │ M │< ,│> .│? /│  Shift   │     │ ↑ │     │ 1 │ 2 │ 3 │   │
 * ├─────┬──┴─┬─┴──┬┴───┴───┴───┴───┴───┴──┬┴───┼───┴┬────┬────┤ ┌───┼───┼───┐ ├───┴───┼───┤ E││
 * │ Ctrl│    │Alt │         Space         │ Alt│    │    │Ctrl│ │ ← │ ↓ │ → │ │   0   │ . │←─┘│
 * └─────┴────┴────┴───────────────────────┴────┴────┴────┴────┘ └───┴───┴───┘ └───────┴───┴───┘
 */

#include<iostream>
#include<cstdio>
#include<cstring>
#include<queue>
#include<vector>
#include<algorithm>
#include<map>
#include<cmath>
using namespace std;
#define inf 0x3f3f3f3f
#define PI acos(-1)
const int mod=1e9+7;
const int M=1e5 + 10;
typedef long long ll;
int a[100];
int main()
{
    int T,n,m,i,j,kk;
    scanf("%d",&T);
    for(kk=1;kk<=T;kk++)
    {
        memset(a,0,sizeof(a));
        int aa,bb,cc,dd,ee;
        scanf("%d%d%d%d%d",&aa,&bb,&cc,&dd,&ee);
        for(i=1;i<=48;i++)
            a[i]=aa;
        for(i=49;i<=56;i++)
            a[i]=bb;
        for(i=57;i<=60;i++)
            a[i]=cc;
        for(i=61;i<=62;i++)
            a[i]=dd;
        a[63]=ee;
        scanf("%d",&n);
        int ans=0,q;
        for(i=1;i<=n;i++)
        {
            scanf("%d",&q);
            ans+=a[q];
        }
        printf("Case #%d: %d0000\n",kk,ans);
    }
    return 0;
}