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.

Example

Sample Input

Sample Output

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
2016-08-17 15:36:04 vaibhav goyal
Question is Just the condition of Simple Linear Diophantine Equation :P
2016-08-15 23:29:03
take care of \n ....get wa a lot..
2016-08-07 01:30:15
note space between : and Yes/No .
2016-07-08 00:50:13
question is wrong
a b and c can be negative ....heres its give a>=1 WTF
got ac taking as -ve too
2016-06-11 12:13:46
easy. :)
2016-06-08 12:50:22
hell to output format.....WA for dat ...take care .... easy othrwise :-)
2016-06-04 12:39:52 Sarthak Munshi
3 lines in python !
2016-04-07 10:12:01
I think you should familiar with diophantine equation to solve this problem :)
2016-03-24 05:12:54
OMG! Java took 3.2s O_o
2016-01-30 19:09:00
good one in basics...take care of output format cost me WA.. :)
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