Codeforces Round #406 (Div. 2) A MONSTER

A. The Monster

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

A monster is chasing after Rick and Morty on another planet. They're so frightened that sometimes they scream. More accurately, Rick screams at times b, b + a, b + 2a, b + 3a, ...and Morty screams at times d, d + c, d + 2c, d + 3c, ....

The Monster will catch them if at any point they scream at the same time, so it wants to know when it will catch them (the first time they scream at the same time) or that they will never scream at the same time.

Input

The first line of input contains two integers a and b (1 ≤ a, b ≤ 100).

The second line contains two integers c and d (1 ≤ c, d ≤ 100).

Output

Print the first time Rick and Morty will scream at the same time, or  - 1 if they will never scream at the same time.

Examples

input

20 2
9 19

output

82

input

2 1
16 12

output

-1
和蓝桥杯凑包子数目是一个类型的题目
先介绍一下扩展欧几里德定理

对于不完全为 0 的整数 a，b，gcd（a，b）表示 a，b 的最大公约数。那么一定存在整

数 x，y 使得 gcd（a，b）=ax+by。

对于这道题目 我们可以整理出 b+x*a=d+y*c    那么就有 ax+(-yc)=d-b

那么可不可达 我们就看d-b能否为gcd(a,c)的倍数了

上代码

 

#include <iostream>
#include <cstdio>
#include <cstring>
#include <vector>
#include <map>
#include <stack>
#include <algorithm>
#include <queue>
#include <string>
#include <cmath>
#include <cstdlib>
using namespace std;
int gcd(int x,int y)
{
    if(y==0) return x;
    else return gcd(y,x%y);
}
// 是否可达的理解
int main()
{
    int a,b,c,d;
    cin>>a>>b>>c>>d;
    int pos=(d-b);
    if(pos%gcd(a,c)!=0)  cout<<-1<<endl;
    else
    {
        while(b!=d)
        {
            if(b < d) b+=a;
            else d+=c;
        }
        cout<<b<<endl;
    }
    return 0;
}