CodeForces 714A

Description

Today an outstanding event is going to happen in the forest — hedgehog Filya will come to his old fried Sonya!

Sonya is an owl and she sleeps during the day and stay awake from minute l₁ to minute r₁ inclusive. Also, during the minute k she prinks and is unavailable for Filya.

Filya works a lot and he plans to visit Sonya from minute l₂ to minute r₂ inclusive.

Calculate the number of minutes they will be able to spend together.

Input

The only line of the input contains integers l₁, r₁, l₂, r₂ and k (1 ≤ l₁, r₁, l₂, r₂, k ≤ 10¹⁸, l₁ ≤ r₁, l₂ ≤ r₂), providing the segments of time for Sonya and Filya and the moment of time when Sonya prinks.

Output

Print one integer — the number of minutes Sonya and Filya will be able to spend together.

Sample Input

Input

1 10 9 20 1

Output

2

Input

1 100 50 200 75

Output

50

Hint

In the first sample, they will be together during minutes 9 and 10.

In the second sample, they will be together from minute 50 to minute 74 and from minute 76 to minute 100.

算是一道非常非常简单的贪心做法题吧。

但我还是WA了无数发，只因为那个被我犯过好几次的错误，不是第一次犯这种错了。

只考虑到了两个区间相交和相离的情况，没有考虑到内含的情况。

这题应该注意的是k是在相交区间断点的问题。

 #include<stdio.h>
 #include<stdlib.h>
 #include<string.h>
 #include<math.h>
 #include<algorithm>
 #include<queue>
 #include<stack>
 #include<deque>
 #include<iostream>
 using namespace std;
 typedef long long  LL;

 struct con{
     LL head;
     LL tail;
 }con[];
 LL cmp(struct con a,struct con b)
 {
     if(a.head==b.head)
         return a.tail<b.tail;
     return a.head<b.head;
 }
 int main()
 {
     LL i,p,j,n;
     LL k,ans=;
 //    scanf("%lld",&p);
 //    printf("%lld\n",p);
     for(i=;i<=;i++)
     {
         scanf("%lld%lld",&con[i].head,&con[i].tail);
     }
     scanf("%lld",&k);
     sort(con,con+,cmp);

     if(con[].head<=con[].tail&&(k>con[].tail||k<con[].head))
     {
         ans=con[].tail-con[].head+;
     }
     if(con[].head<=con[].tail&&(k<=con[].tail&&k>=con[].head))
     {
         ans=con[].tail-con[].head;
     }
     if(con[].tail>=con[].tail&&(k>con[].tail||k<con[].head))
     {
         ans=con[].tail-con[].head+;
     }
     if(con[].tail>=con[].tail&&(k<=con[].tail&&k>=con[].head))
     {
         ans=con[].tail-con[].head;
     }
     printf("%lld\n",ans);
     return ;
 }