Codeforces Round #196 (Div. 2) B. Routine Problem

B. Routine Problem

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Manao has a monitor. The screen of the monitor has horizontal to vertical length ratio a:b. Now he is going to watch a movie. The movie's frame has horizontal to vertical length ratio c:d. Manao adjusts the view in such a way that the movie preserves the original frame ratio, but also occupies as much space on the screen as possible and fits within it completely. Thus, he may have to zoom the movie in or out, but Manao will always change the frame proportionally in both dimensions.

Calculate the ratio of empty screen (the part of the screen not occupied by the movie) to the total screen size. Print the answer as an irreducible fraction p / q.

Input

A single line contains four space-separated integers a, b, c, d (1 ≤ a, b, c, d ≤ 1000).

Output

Print the answer to the problem as "p/q", where p is a non-negative integer, q is a positive integer and numbers p and q don't have a common divisor larger than 1.

Sample test(s)

input

1 1 3 2

output

1/3

input

4 3 2 2

output

1/4

Note

Sample 1. Manao's monitor has a square screen. The movie has 3:2 horizontal to vertical length ratio. Obviously, the movie occupies most of the screen if the width of the picture coincides with the width of the screen. In this case, only 2/3 of the monitor will project the movie in the horizontal dimension:

Sample 2. This time the monitor's width is 4/3 times larger than its height and the movie's frame is square. In this case, the picture must take up the whole monitor in the vertical dimension and only 3/4 in the horizontal dimension:

#include <iostream>#include <stdio.h>#include <string.h>using namespace std;int gcd(int a,int b){    if(a==0)    return b;    return gcd(b%a,a);}int main(){   int a1,a2,a3,a4,l1,l2,l3;   //printf("%d",gcd(2,4));   while(scanf("%d%d%d%d",&a1,&a2,&a3,&a4)!=EOF)   {       l1=a1*a4,l2=a2*a3;       if(l1>l2)       {           l2=l1-l2;           l3=gcd(l1,l2);           l1/=l3,l2/=l3;           printf("%d/%d\n",l2,l1);       }       else if(l1==l2)       {           printf("0/1\n");       }       else if(l1<l2)       {           l1=l2-l1;           l3=gcd(l1,l2);           l1/=l3,l2/=l3;           printf("%d/%d\n",l1,l2);       }   }    return 0;}