CF

1.题目描述：

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

31 52 62 322 46 8

output

3

input

31 52 63 722 41 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.

2.题意概述：

一个人有编程课和象棋课，至少上一节，象棋课给你若干的选择上课时间段选择，编程课给你若干个上课时间段选择。这个人想获得尽可能多的休息时间，问你最长的休息时间。（定义两门课的休息时间是指早开始那门课的结束时间到晚开始那门课的开始时间）。

3.解题思路：

一开始往排序方向想去了，其实不然，直接记录一下象棋课和编程课分别的最早结束时间和最晚开始时间就行，最优解就是这个。如果出现负数说明最优情况也是区间重叠，那么这时候休息时间就是0。

4.AC代码：

#include <bits/stdc++.h>#define INF 0x3f3f3f3f#define maxn 200200#define N 1111#define eps 1e-6#define pi acos(-1.0)#define e exp(1.0)using namespace std;const int mod = 1e9 + 7;typedef long long ll;typedef unsigned long long ull;int main(){#ifndef ONLINE_JUDGEfreopen("in.txt", "r", stdin);freopen("out.txt", "w", stdout);long _begin_time = clock();#endifint n, m;while (~scanf("%d", &n)){int a_l = 0, a_r = INF, b_l = 0, b_r = INF;for (int i = 0; i < n; i++){int x, y;scanf("%d%d", &x, &y);a_l = max(a_l, x);a_r = min(a_r, y);}scanf("%d", &m);for (int i = 0; i < m; i++){int x, y;scanf("%d%d", &x, &y);b_l = max(b_l, x);b_r = min(b_r, y);}int ans = max(a_l - b_r, b_l - a_r);printf("%d\n", max(ans, 0));}#ifndef ONLINE_JUDGElong _end_time = clock();printf("time = %ld ms.", _end_time - _begin_time);#endifreturn 0;}