CF787A：The Monster（扩展欧几里得）

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

Note

In the first sample testcase, Rick's 5th scream and Morty's 8th time are at time 82.

In the second sample testcase, all Rick's screams will be at odd times and Morty's will be at even times, so they will never scream at the same time.

题意：给出a,b,c,d四个数，两个人在尖叫，第一个人在b,b+a,b+2a,b+3a......时刻尖叫，第二个人在d,d+c,d+2c,d+3c......时刻尖叫，问他们第一次在相同时刻尖叫是什么时候。

思路：设第一个人x，第二个y，显然有b+ax = d+cy，ax-cy=d-b，求最小的x解即可，由于y又要保证为非负数，x要一直增加到y为非负数。

# include <bits/stdc++.h>using namespace std;int a, b, c, d, x, y, gd;int ex_gcd(int a, int b){    if(b == 0)    {        x = 1;        y = 0;        return a;    }    gd = ex_gcd(b, a%b);    int t = x;    x = y;    y = t-(a/b)*y;    return gd;}int main(){    int A, B, C, D, t, tmp=0;    scanf("%d%d%d%d",&A,&B,&C,&D);    a = A;    b = -C;    c = D-B;    gd = ex_gcd(a, b);    if(c % gd != 0)        return 0*puts("-1");    x = x*c/gd;    y = y*c/gd;    int k = b/gd;    x = (x%k+k)%k;    if(x < 0) x += abs(k);    /*    while((c - a*x)/b < 0) x += abs(k);    下面的东西可以用这个暴力循环代替。    */    if((c - a*x)/b < 0)    {        tmp = a*x-c;        k = abs(k);        t = (tmp%(a*k)+(a*k))%(a*k);        if(t < 0)            t += a*k;        tmp = (t-tmp)/(a*k);    }    printf("%d\n",B+A*(x+k*tmp));    return 0;}