# DCP-255: Parand()thesis Back to All Problems

Hard Math > Counting

![Tom and Jerry][1] Tom and Jerry are two famous programmers. Jerry practices hard all day long, but on the other hand Tom always tries to find shortcuts. We all know there are no shortcuts to success. But somehow until this day, Tom managed to be successful. This misjudgment demotivates Jerry to concentrate his programming practice. That’s why, he finally makes a decision to reveal the true self of Tom. Jerry challenges Tom to write a code that generates valid parenthesis expressions. Rather than writing a good code, Tom copies a code from the internet and shows to Jerry. Unfortunately, that code generates random parenthesis expressions without checking any validity at all. Tom’s code takes a number **N** and generates an expression of parenthesis consist of **N pairs** randomly. For example, some of the randomly generated **2 pairs** parenthesis expression can be **((((, ()((, ))(), (()),** … etc. As you can see some of them are valid and others are not. Now, the critical moment has come and Jerry will give the input **N** to Tom’s code. But he fears, what if he loses the challenge? But as a good programmer, you know that Jerry needs to be afraid a little, right? So, as a helping hand, you will calculate the probability of winning the challenge by Tom for a given value of **N** and show it to Jerry that how poor Tom’s code is. This will enhance the confidence of Jerry of course! **More specifically,** what is the probability of getting a valid **N pairs** parenthesis expression generated randomly. This probability can be written as an **irreducible** fraction **A / B**. Your task is to find the values of **A** and **B** (Jerry does not like to deal with floating point numbers). If **V** is a valid configuration then, **VV** and **(V)** are also valid. For example, **V=()**. Input: ------ Input starts with an integer **T (<=10)**, denoting the number of test cases. Each case contains an integer **N (1 ≤ N ≤ 10000)** denoting the **number of parenthesis pairs** in the expression. Output: ------- For each case, print the case number and the desired probability. See the samples for exact formatting. If the desired probability is **P**, then print two numbers **A** and **B** where **P = A / B** and **A** and **B** are coprime **(A >= 0, B >= 1, gcd(A, B) = 1)**. Since these numbers may be too large, print them modulo **10^9 + 7**. Note that **A** and **B** must be coprime before their remainders modulo **10^9 + 7** are taken. Sample Input ------------ 2 1 2 Sample Output ------------- Case 1: 1 4 Case 2: 1 8 [1]: https://s3-ap-southeast-1.amazonaws.com/devskillimagestorage/questionimages/513575f6-8391-c79f-010f-08d44ec324d4_368d4cf0016d4cfaa53725e3fd345d89_W650xH487.jpg

### Problem Limits

 Language Time Limit (seconds) C 1.00 C++ 1.00 C++14 1.00 C# 1.00 Go 1.00 Java 1.00 JavaScript 1.00 Objective-C 1.00 Perl 1.00 PHP 1.00 Python 1.00 Python3 1.00 Ruby 1.00 VB.Net 1.00

# 30/93

Solve/Submission

### Ranking

# User Language Timing
01 liar Cpp14 0.00s
02 sayedgkm Cpp14 0.00s
03 PKP_007 Cpp14 0.00s
04 nasif2587 Cpp14 0.00s
05 feodorv Cpp14 0.00s
06 Robbinb1993 Cpp14 0.00s
07 AlirezaNa Cpp14 0.00s
08 SleepyBrain Cpp14 0.01s
09 akazad_cse13_ruet Cpp14 0.01s
10 mahbubcseju Cpp14 0.01s
11 Zeronfinity Cpp14 0.01s
12 sumit1993 Cpp14 0.01s
13 Dariwala Cpp14 0.01s
14 dmehrab06 Cpp14 0.02s
15 habib_rahman Cpp14 0.02s
16 as_couple Cpp14 0.02s
17 shakil_ruet Cpp14 0.05s
18 ssavi Cpp14 0.09s
19 Morass Cpp14 0.12s
20 7Mahfuz Cpp14 0.14s
21 I_See_You Cpp14 0.14s
22 sahedsohel Cpp14 0.16s
23 tariqiitju Cpp14 0.26s
24 seyedssz Cpp14 0.37s
25 MRoy Cpp14 0.61s
Feedback