## 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 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 "<