Manthan, Codefest 16 -A Ebony and Ivory

time limit per test	2 seconds
memory limit per test	256 megabytes
input standard	input
output standard	output

Dante is engaged in a fight with 「The Savior」. Before he can fight it with his sword, he needs to break its shields. He has two guns, Ebony and Ivory, each of them is able to perform any non-negative number of shots.

For every bullet that hits the shield, Ebony deals a units of damage while Ivory deals b units of damage. In order to break the shield Dante has to deal exactly c units of damage. Find out if this is possible.

Input

The first line of the input contains three integers a, b, c (1 ≤ a, b ≤ 100, 1 ≤ c ≤ 10 000) — the number of units of damage dealt by Ebony gun and Ivory gun, and the total number of damage required to break the shield, respectively.

Output

Print 「Yes」 (without quotes) if Dante can deal exactly c damage to the shield and 「No」 (without quotes) otherwise.

Examples

input
4 6 15
output
No
input
3 2 7
output
Yes
input
6 11 6
output
Yes

Note

In the second sample, Dante can fire 1 bullet from Ebony and 2 from Ivory to deal exactly 1·3 + 2·2 = 7 damage. In the third sample, Dante can fire 1 bullet from ebony and no bullets from ivory to do 1·6 + 0·11 = 6 damage.

題意：判斷c 能不能由A,B組成
暴力枚舉便可

#include <cstdio>
#include <cstring>
#include <cmath>
#include <cstdlib>
#include <string>
#include <stack>
#include <queue>
#include <vector>
#include <set>
#include <list>
#include <map>
#include <iostream>
#include <algorithm>

using namespace std;

typedef long long LL;

const int INF = 0x3f3f3f3f;

const double eps = 1e-6;

const double PI = acos(-1.0);

int a,b,c;

int main()
{
    cin>>a>>b>>c;

    bool flag=false;
    for(int i=0;a*i<=c;i++)
    {
        if((c-a*i)%b==0)
        {
            flag= true;

            break;
        }
    }

    if(flag)
    {
        cout<<"Yes"<<endl;
    }
    else

    {
        cout<<"No"<<endl;
    }
    return 0;
}