## ARBITRAG - Arbitrage

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Arbitrage is the use of discrepancies in currency exchange rates to transform one unit of a currency into more than one unit of the same currency. For example, suppose that 1 US Dollar buys 0.5 British pounds, 1 British pound buys 10.0 French francs, and 1 French franc buys 0.21 US dollars. Then, by converting currencies, a clever trader can start with 1 US dollar and buy 0.5 * 10.0 * 0.21 = 1.05 US dollars, making a profit of 5 percent.

Your job is to write a program that takes a list of currency exchange rates as input and then determines whether arbitrage is possible or not.

### Input

The input file will contain one or more test cases. On the first line of each test case there is an integer n (1<=n<=30), representing the number of different currencies. The next n lines each contain the name of one currency. Within a name no spaces will appear. The next line contains one integer m, representing the length of the table to follow. The last m lines each contain the name ci of a source currency, a real number rij which represents the exchange rate from ci to cj and a name cj of the destination currency. Note that ci and cj may be the same currency. Exchanges which do not appear in the table are impossible.
Test cases are separated from each other by a blank line. Input is terminated by a value of zero (0) for n.

### Output

For each test case, print one line telling whether arbitrage is possible or not in the format "Case case: Yes", respectively "Case case: No".

### Example

```Input:
3
USDollar
BritishPound
FrenchFranc

3
USDollar 0.5 BritishPound
BritishPound 10.0 FrenchFranc
FrenchFranc 0.21 USDollar

3
USDollar
BritishPound
FrenchFranc

6
USDollar 0.5 BritishPound
USDollar 4.9 FrenchFranc
BritishPound 10.0 FrenchFranc
BritishPound 1.99 USDollar
FrenchFranc 0.09 BritishPound
FrenchFranc 0.19 USDollar

0

Output:
Case 1: Yes
Case 2: No
```

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 < Previous 1 2 3 Next > shohan2032: 2023-10-10 14:30:26 can any body give me the bellman ford solution for this problem??Though i have solved it with warshall. ive1010: 2022-03-22 16:34:12 Bellman ford with time complexity : O(N^4) vineetjai: 2020-07-24 14:37:20 Don't think of taking log or finding cycles and don't try to convert it into bellman ford. Use Floyd Warshall for a quick answer. roopammishra: 2020-03-23 22:55:54 Wait ! What !.. I made some random changes and got AC :) enigmus: 2019-06-06 10:56:26 Printing uppercase answers costed me a couple of WAs :(. Solvable by basic bellman-ford, taking negative logarithm of edges is not required, just follow the edges by multiplying abhinav_jain02: 2019-05-23 14:28:31 AC in one go Apply bellman ford for every weakly connected component surajkumar_cse: 2019-01-15 23:13:02 The problem demands to product the weight of the edges ,,, after lots of thinking and online help ... I was able to convert the product into sum. Hint: 12th standard maths. jmr99: 2018-08-15 15:23:05 FloyWar!! abdelhameedddd: 2018-08-07 10:35:12 AC USING FLOYD :) anirudnits: 2018-06-18 15:04:47 @siva2697 thanks and my first AC using Floyd warshall :)

 Added by: Vincenzo Bonifaci Date: 2011-08-07 Time limit: 1s Source limit: 50000B Memory limit: 1536MB Cluster: Cube (Intel G860) Languages: All except: ASM64 Resource: University of Ulm Local Contest 1996