CEQU - Crucial Equation

Let us see the following equation,

                                            ax+by=c

Given three positive integers a, b and c. You have to determine whether there exists at least one solution for some integers value of x and y where x, y may be negative or non-negative integers.

For example if a=2, b=4 and c=8 then the equation will be 2x+4y=8, and hence, for x=2 and y=1, there exists a solution.

Let us see another example for a=3, b=6 and c=7, so the equation will become 3x+6y=7 and there exists no solution satisfying this equation.

Input

Input starts with an integer T (1<=T<=105) denoting the number of test cases. Each test case contains three integers a, b, and c. (1<=a, b, c<=106).

Output

For each test case of input print the case number and “Yes” if there exists at least one solution, print “No” otherwise.

Sample Input

Output for Sample Input

2
2 4 8
3 6 7

Case 1: Yes
Case 2: No

Problem Setter: Md Abdul Alim, Dept. of Computer Science, Bangladesh University of Business & Technology


Added by:Alim
Date:2014-10-15
Time limit:3s
Source limit:50000B
Memory limit:1536MB
Cluster: Cube (Intel G860)
Languages:All except: ASM64 GOSU
Resource:Own Problem

hide comments
2017-04-29 15:16:02
there is only two cases case 1 and case 2 or more,and space between case and 1 or not.
2017-04-29 15:13:27
i didn't understood input/output format.
2017-04-11 18:41:20
very easy AC in one go...!
2017-03-31 16:36:47
Gcd is enough.
2017-02-18 19:33:57
space between : and Yes / No (*it cause me two WA)
2017-02-14 22:08:14
simple one for who knows diophantine equation...
2017-01-07 14:19:01
easy :) my 20th..got wa due to output format.Use Diophantine Equation...hope this helps:)
2016-12-02 15:43:08
Linear Diophantine Equations condition
2016-11-26 05:11:31
trivial
2016-11-25 20:15:01
You don't have to just print "Yes" and "No". Literally, you gotta print
Case 1: ...
Case 2: ...
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