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
2018-06-08 07:51:06
Take care of how to print the output!
2017-08-12 07:50:25
AC in one go!........
2017-07-14 14:25:41
Bézout's identity :)
2017-07-11 07:51:49
AC in one go. Easy one
2017-07-02 10:21:11
My nth..
AC in one go :D
2017-06-29 14:23:19
how to print output
2017-05-29 21:22:50
AC in one go :)
My 50th !! :D
2017-05-23 07:19:29
Don't forget to add space between : and "yes" or "no"
2017-04-29 15:16:02
there is only two cases case 1 and case 2 or more,and space between case and 1 or not.
2017-04-29 15:13:27
i didn't understood input/output format.
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