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 nonnegative 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<=10^{5}) denoting the number of test cases. Each test case contains three integers a, b, and c. (1<=a, b, c<=10^{6}).
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 
Case 1: Yes 
Problem Setter: Md Abdul Alim, CEO and Founder at CodeMask
hide comments
sankalp_7:
20210624 09:17:07
Last edit: 20210709 05:23:39 

itsabi_z1:
20210622 06:05:01
pay attention to the output format. 

anchord:
20210607 06:30:31
Good question! only about the basic concept about linear diophantine, pretty simple 

messier22:
20210508 13:35:46
I'm a faggot. Last edit: 20210508 22:53:03 

crawler_123:
20210423 20:01:27
The Problem is based on finding solution for Linear Diophantine Equation, hope this hint will help any one of you. 

sk128:
20210405 15:34:14
Got AC ?


emily50:
20210316 17:29:54
@abu_rifat ans is ok ,but what is the logic? 

mantavya:
20200722 16:15:13
Output it in cout<<"Case "<<i<<": Yes"<<'\n'; this format. :) 

amit_dubey99:
20200616 19:04:34
those Case 1 2 stuff sucks, cost me WA, but m happy finally it got AC :) 

priyankarai:
20200609 16:21:31
very bad problem, got WA just because wrong display of output,no focus on concept but only focussing on visuals 
Added by:  Md Abdul Alim 
Date:  20141015 
Time limit:  3s 
Source limit:  50000B 
Memory limit:  1536MB 
Cluster:  Cube (Intel G860) 
Languages:  All except: ASM64 GOSU 
Resource:  Own Problem 