Anton and Classes（贪心）

B. Anton and Classes

time limit per test

4 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Anton likes to play chess. Also he likes to do programming. No wonder that he decided to attend chess classes and programming classes.

Anton has n variants when he will attend chess classes, i-th variant is given by a period of time (l1, i, r1, i). Also he has m variants when he will attend programming classes, i-th variant is given by a period of time (l2, i, r2, i).

Anton needs to choose exactly one of n possible periods of time when he will attend chess classes and exactly one of m possible periods of time when he will attend programming classes. He wants to have a rest between classes, so from all the possible pairs of the periods he wants to choose the one where the distance between the periods is maximal.

The distance between periods (l1, r1) and (l2, r2) is the minimal possible distance between a point in the first period and a point in the second period, that is the minimal possible |i - j|, where l1 ≤ i ≤ r1 and l2 ≤ j ≤ r2. In particular, when the periods intersect, the distance between them is 0.

Anton wants to know how much time his rest between the classes will last in the best case. Help Anton and find this number!

Input

The first line of the input contains a single integer n (1 ≤ n ≤ 200 000) — the number of time periods when Anton can attend chess classes.

Each of the following n lines of the input contains two integers l1, i and r1, i (1 ≤ l1, i ≤ r1, i ≤ 109) — the i-th variant of a period of time when Anton can attend chess classes.

The following line of the input contains a single integer m (1 ≤ m ≤ 200 000) — the number of time periods when Anton can attend programming classes.

Each of the following m lines of the input contains two integers l2, i and r2, i (1 ≤ l2, i ≤ r2, i ≤ 109) — the i-th variant of a period of time when Anton can attend programming classes.

Output

Output one integer — the maximal possible distance between time periods.

Examples

input

Copy

3
1 5
2 6
2 3
2
2 4
6 8

output

3

input

Copy

3
1 5
2 6
3 7
2
2 4
1 4

output

0

Note

In the first sample Anton can attend chess classes in the period (2, 3) and attend programming classes in the period (6, 8). It's not hard to see that in this case the distance between the periods will be equal to 3.

In the second sample if he chooses any pair of periods, they will intersect. So the answer is 0.

题意：a课程的可选时间段为n个，b课程的可选时间段有m个，选择a和b中的时间段个一个，使两个课程之间的时间间隔最大，如果时间段重叠，则时间间隔为0.

思路：简单贪心，按照a课程结束时间由大到小排列，b课程的课程开始时间由小到大排列用边最大的减去前边最小的，得到temp1，之后ab调换，在得到第二个差值temp2，在两个差值都大于0的前提下比较保留最大的，否则时间差定为0.

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
struct node
{
    int s,e;
}a[200005],b[200005];
int cmpe(node s1,node s2)
{
    return s1.e>s2.e;///降序
}
int cmps(node h1,node h2)
{
    return h1.s<h2.s;///升序
}
int main()
{
    int x,y;
    scanf("%d",&x);
    for(int i=0;i<x;i++)
        scanf("%d %d",&a[i].s,&a[i].e);
    scanf("%d",&y);
    for(int i=0;i<y;i++)
        scanf("%d %d",&b[i].s,&b[i].e);
    sort(a,a+x,cmpe);
    sort(b,b+y,cmps);
   /* for(int i=0;i<x;i++)
        printf("%d %d\n",a[i].s,a[i].e);
    for(int i=0;i<y;i++)
        printf("%d %d\n",b[i].s,b[i].e);*/
    int Max=-1,temp1,temp2;
    temp1=b[y-1].s-a[x-1].e;
    sort(b,b+y,cmpe);
    sort(a,a+x,cmps);
    /*for(int i=0;i<x;i++)
        printf("*%d %d\n",a[i].s,a[i].e);
    for(int i=0;i<y;i++)
        printf("*%d %d\n",b[i].s,b[i].e);*/
    temp2=a[x-1].s-b[y-1].e;
    if(temp1>temp2&&temp1>0)
        Max=temp1;
    else if(temp1<=temp2&&temp2>0)
        Max=temp2;
    else
        Max=0;
    printf("%d\n",Max);
        return 0;
}