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
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: ...
2016-11-13 17:07:03
a=3 b=7 c=11 AC solution giving Yes. weird.
2016-10-27 08:55:42 E Naveen Kumar
AC in a GO...!!
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