Problem: A linguist, Nodvic Natharus Damenhof (commonly called Dr. Usoperant), invented an artificial language Usoperant in 2007. The word usoperant means ‘one which tires’. Damenhof’s goal was to create a complex and pedantic language that would remind many difficulties in universal communications. Talking in Usoperant, you should remember the importance of choosing your words in many conversations.

For example of the complexity, you may be confused by the way to say large numbers. Usoperant has some words that indicate exponential numbers in decimal, described as 10p where p is a positive integer. In terms of English, those words might include thousand for 103 (1,000), million for 106 (1,000,000), and undecillion for 1036 (1,000,000,000,000,000,000,000,000,000,000,000,000).

You can concatinate those words to express larger numbers. When two words w1 and w2 mean numbers 10p1 and 10p2 respectively, a concatinated word w1w2 means 10p1+p2. Using the above examples in English (actually the following examples are incorrect in English), you can say 109 by millionthousand, 1012 by millionmillion and 1039 by undecillionthousand. Note that there can be multiple different expressions for a certain number. For example, 109 can also be expressed by thousandthousandthousand. It is also possible to insert separators between the adjacent components like million-thousand and thousand-thousand-thousand.

In this problem, you are given a few of such words, their representing digits and an expression of a certain number in Usoperant. Your task is to write a program to calculate the length of the shortest expression which represents the same number as the given expression.

The expressions in the input do not contain any separators like millionthousand. In case of ambiguity, the expressions should be interpreted as the largest number among possibles. The resultant expressions should always contain separators like million-thousand, so we can distinguish, for example, x-x (a concatinated word of two x's) and xx (just a single word). The separators should not be counted in the length.

Note:

In the fourth case, the word abcdefgh can be separated as abcd-efgh (representing 103 ) and abcde-fgh (representing 1010 ), and by the rule described in the problem statement, this word should be interpreted as 1010. As this number can be expressed by yy, the length of the shortest expression is two (as in the sample output). The existence of y-y (representing 1016 ) is not a problem here, because we can distinguish yy and y-y as long as we always use separators.



Input

The input consists of multiple test cases. Each test case begins with a line including an integer N (1 ≤ N ≤ 100). The integer N indicates the number of words in the dictionary of exponential numbers.

The following N lines give the words in the dictionary. Each line contains a word wi and an integer pi (1 ≤ i ≤ N, 1 ≤ pi ≤ 10), specifying that the word wi expresses an exponential number 10pi in Usoperant. Each wi consists of at most 100 alphabetical letters.

Then a line which contains an expression of a number in Usoperant follows. The expression consists of at most 200 alphabetical letters.

The end of the input is indicated by a line which contains only a single 0.

Output

For each test case, print a line which contains the case number and the length of the shortest expression which represents the same number as the given expression in the input.

Example

Input

3
billion 9
million 6
thousand 3
billionmillionthousand
3
oku 8
sen 3
man 4
okusenman
2
nico 2
video 4
niconicovideo
6
abcd 1
efgh 2
abcde 4
fgh 6
y 8
yy 10
abcdefgh
0


Output

Case 1: 14
Case 2: 9
Case 3: 10
Case 4: 2

Solution (in CPP):

Solution:
//
//  main.cpp
//  160324
//
//  Created by ?加寿 on 16/3/24.
//  Copyright ? 2016年 chenhuan001. All rights reserved.
//

#include <iostream>
#include <stdio.h>
#include <string.h>
#include <string>
#include <map>
#include <algorithm>
using namespace std;

#define N 110
map<string,int> cao;

string saves[N];
int savek[N];
int main(int argc, const char * argv[]) {
    int n;
    int tt=1;
    while(cin>>n&&n)
    {
        cao.clear();
        for(int i=0;i<n;i++)
        {
            string tmp;
            int key;
            cin>>tmp>>key;
            saves[i] = tmp;
            savek[i] = key;
            cao[tmp] = max(cao[tmp],key);
        }
        //第一个背包
        char g[222];
        scanf("%s",g);
        int len=strlen(g);
        int dp[22200];
        memset(dp,-1,sizeof(dp));
        dp[0]=0;
        for(int i=0;i<len;i++)
        {
            //?个地方有点??
            //dp[i+1] = max(dp[i+1],dp[i]);
            if(dp[i]==-1)
            {
                continue;
            }
            string str="";
            for(int j=i;j<len;j++)
            {
                str+=g[j];
                if(cao[str] != 0)
                {
                    dp[j+1] = max(dp[j+1],dp[i]+cao[str]);
                }
            }
        }
        int mx=dp[len];
        //再用2000的背包.
        for(int i=0;i<2002;i++)
            dp[i] = 1e6;
        dp[0] = 0;
        //无限背包
        for(int i=0;i<mx;i++)
        {
            for(int j=0;j<n;j++)
            {
                int sz=saves[j].length();
                dp[ i+savek[j] ] = min(dp[i+savek[j]],dp[i]+sz);
            }
        }
        printf("Case %d: ",tt++);
        cout<<dp[mx]<<endl;
    }
    return 0;
}