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[10];
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=0;
//    scanf("%lld",&p);
//    printf("%lld
",p);
    for(i=0;i<=1;i++)
    {
        scanf("%lld%lld",&con[i].head,&con[i].tail);
    }
    scanf("%lld",&k);
    sort(con,con+2,cmp);

    if(con[1].head<=con[0].tail&&(k>con[0].tail||k<con[1].head))
    {
        ans=con[0].tail-con[1].head+1;
    }
    if(con[1].head<=con[0].tail&&(k<=con[0].tail&&k>=con[1].head))
    {
        ans=con[0].tail-con[1].head;
    }
    if(con[0].tail>=con[1].tail&&(k>con[1].tail||k<con[1].head))
    {
        ans=con[1].tail-con[1].head+1;
    }
    if(con[0].tail>=con[1].tail&&(k<=con[1].tail&&k>=con[1].head))
    {
        ans=con[1].tail-con[1].head;
    }
    printf("%lld
",ans);
    return 0;
}