codeforce 787 A. The 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 29 19

output

82

input

2 116 12

output

-1

这个题目使我对于扩展欧几里得的理解更深了。

b+x*a=d+y*c -->a*x-c*y=d-b 因为适用条件是a、c为不完全为零的非负整数。

所以这个题目潜在的要求是：在多组解中选出x>0，y<0的情况。

具体细节详见代码注释；

#include<stdio.h>int ans;int Exgcd(int a,int b,int &x,int &y){if(b==0){x=1; y=0;return a;}int gg=Exgcd(b,a%b,y,x);y-=a/b*x;return gg;}int cal(int a,int b,int c,int d){int x,y;int gg=Exgcd(a,c,x,y);if((d-b)%gg!=0)return 0;else{int m=c/gg;printf("xy = %d%d",x,y);x=(d-b)/gg*x;//先按照gg与d-b倍数关系转化 y=(d-b)/gg*y;x=((x%m)+m)%m;//将x转化为最小正整数 y=(d-b-a*x)/c;while(y>0)//在y大于0的时候不断增大x。 {x+=m;//在通解中找出特解。 y=(d-b-a*x)/c;}ans=b+a*x;}return 1;}int main(){int a,b,c,d;while(~scanf("%d%d%d%d",&a,&b,&c,&d)){if(cal(a,b,c,d))printf("%d\n",ans);elseprintf("-1\n");}return 0;}