Problem:
Score : 200 points

Problem Statement

Alice and Bob are controlling a robot. They each have one switch that controls the robot.

Alice started holding down her button A second after the start-up of the robot, and released her button B second after the start-up.

Bob started holding down his button C second after the start-up, and released his button D second after the start-up.

For how many seconds both Alice and Bob were holding down their buttons?

Constraints

0≤A<B≤100

0≤C<D≤100

All input values are integers.

Input

Input is given from Standard Input in the following format:

A B C D

Output

Print the length of the duration (in seconds) in which both Alice and Bob were holding down their buttons.

Sample Input 1

0 75 25 100

Sample Output 1

50

Alice started holding down her button 0 second after the start-up of the robot, and released her button 75 second after the start-up.

Bob started holding down his button 25 second after the start-up, and released his button 100 second after the start-up.

Therefore, the time when both of them were holding down their buttons, is the 50 seconds from 25 seconds after the start-up to 75 seconds after the start-up.

Sample Input 2

0 33 66 99

Sample Output 2

0

Alice and Bob were not holding their buttons at the same time, so the answer is zero seconds.

Sample Input 3

10 90 20 80

Sample Output 3

60

Solution in C (GCC 5.4.1) (judge verdict: Accepted)

#include<stdio.h>
int main()
{
	int a,b,c,d;
	while (scanf("%d%d%d%d",&a,&b,&c,&d)!=EOF)
	{
		if(c>=b||a>=d)
		{
			printf("0\n");
		}
		if(a<c&&c<b&&b<d)
		{
			printf("%d\n",b-c);
		}
		if(c<a&&a<b&&d<b&&d>a)
		{
			printf("%d\n",d-a);
		}
		if(c>a&&b>d)
		{
			printf("%d\n",d-c);
		}
		if(a>c&&d>b)
		{
			printf("%d\n",b-a);
		}
		if(a==c&&b==d)
		{
			printf("%d\n",b-a);
		}
		if(a==c&&b>d)
		{
			printf("%d\n",d-c);
		}
		if(b==d&&a<c)
		{
			printf("%d\n",d-c);
		}
		if(a==c&&d>b)
		{
			printf("%d\n",b-a);
		}
		if(b==d&&c<a)
		{
			printf("%d\n",b-a);
		}
	}
	return 0;
}