Problem:
Score : 200 points

Problem Statement

We ask you to select some number of positive integers, and calculate the sum of them.

It is allowed to select as many integers as you like, and as large integers as you wish.
You have to follow these, however: each selected integer needs to be a multiple of A, and you need to select at least one integer.

Your objective is to make the sum congruent to C modulo B.
Determine whether this is possible.

If the objective is achievable, print YES. Otherwise, print NO.

Constraints

1 ≤ A ≤ 100

1 ≤ B ≤ 100

0 ≤ C < B

Input

Input is given from Standard Input in the following format:

A B C

Output

Print YES or NO.

Sample Input 1

7 5 1

Sample Output 1

YES

For example, if you select 7 and 14, the sum 21 is congruent to 1 modulo 5.

Sample Input 2

2 2 1

Sample Output 2

NO

The sum of even numbers, no matter how many, is never odd.

Sample Input 3

1 100 97

Sample Output 3

YES

You can select 97, since you may select multiples of 1, that is, all integers.

Sample Input 4

40 98 58

Sample Output 4

YES

Sample Input 5

77 42 36

Sample Output 5

NO

Solution in Java8 (OpenJDK 1.8.0) (judge verdict: Time Limit Exceeded)

import java.util.Scanner;

public class Main {

	public static void main(String[] args) {

		Scanner sc = new Scanner(System.in);

		int A = sc.nextInt();
		int B = sc.nextInt();
		int C = sc.nextInt();
		int i = 1;
		int flag = 0;

		//Aの倍数をBで割った余りを求める
		loop: while (flag == 0) {
			int tmp_rest = A * i % B;
			//もしAの倍数をBで割った余りがCと等しければ、flagをYES用の1にして処理を終了
			if (tmp_rest == C) {
				flag = 1;
				break loop;
			}

			//もしAの倍数をBで割った余りがCと等しくなければ、

			else {
				int j = 1;
				while(j<i) {
					//A * i までの余りを算出して、今回の余りと合致するか調べる
					int tmp_rest2 = A * j % B;
					if (tmp_rest == tmp_rest2) {
						flag = -1;
						break loop;
					}
					else {
						flag = 0;
					}
				}
			}
			//もし重複がなければ、次の処理に移行する
			i++;
		}

		if (flag > 0) {
			System.out.println("YES");
		}
		else {
			System.out.println("NO");
		}

	}
}