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 C++14 (GCC 5.4.1) (judge verdict: Accepted)

#include<iostream>
using namespace std;

int main() {
	int A, B, C;
	cin >> A >> B >> C;
	for (int i = A;i <= A * B;i += A) {
		if (i % B == C) { // 余りがCになるAの倍数が存在した
			cout << "YES";
			return 0;
		}
	} // 余りがCになるAの倍数が無かった
	cout << "NO";
	return 0;
}
/*
31536000のコメント解説欄
ここテンプレで用意してるから、A問題とかだとこの先空欄の危険あり

今、ある数xを任意のAの倍数の総和であるとする
その時、xはAの倍数である
よってそれは1個のAの倍数を用いて表すことができる
つまり、総和をBで割った余りなぞ使わなくても、あるAの倍数をBで割った余りがCであれば良い
また、x>=ABであった時、x-ABをBで割った余りはxをBで割った余りに等しい
よって、A<=x<=ABであると言える
後は全部検証すればすぐ終わる、O(B)
*/