## 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 83 6 7 Case 1: Yes Case 2: No

Problem Setter: Md Abdul Alim, CEO and Founder at CodeMask

 < Previous 1 2 3 4 5 6 7 8 Next > sankalp_7: 2021-06-24 09:17:07 Last edit: 2021-07-09 05:23:39 itsabi_z1: 2021-06-22 06:05:01 pay attention to the output format. anchord: 2021-06-07 06:30:31 Good question! only about the basic concept about linear diophantine, pretty simple messier22: 2021-05-08 13:35:46 I'm a faggot. Last edit: 2021-05-08 22:53:03 crawler_123: 2021-04-23 20:01:27 The Problem is based on finding solution for Linear Diophantine Equation, hope this hint will help any one of you. sk128: 2021-04-05 15:34:14 Got AC ? "Yes" emily50: 2021-03-16 17:29:54 @abu_rifat ans is ok ,but what is the logic? mantavya: 2020-07-22 16:15:13 Output it in cout<<"Case "<