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
2020-01-16 17:25:32
Diophantine equation-hint
2019-12-19 18:14:18
AC in one go!!!!!!!!
2019-12-13 08:13:21
follow the output format exactly costed me 2 wa..
2019-08-26 13:56:30
Hint is...
Find gcd of a and b..
If c%gcd ==0
Then yes otherwise no
2019-06-01 01:21:44
AC in one go
2019-04-15 12:16:02
Useful video for this question : https://www.youtube.com/watch?v=OrVWAYonFIU
2018-08-14 18:58:41
Use euclidean gcd!!
2018-08-14 18:57:55
AC in one go...!!!
2018-06-24 13:28:37
the basic-most question of diophantine equation.
2018-06-20 19:46:39
follow the oupt frmt!!
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