G - Absolute Game

G - Absolute Game

Alice and Bob are playing a game. Alice has an array aa of nn integers, Bob has an array bb of nn integers. In each turn, a player removes one element of his array. Players take turns alternately. Alice goes first.

The game ends when both arrays contain exactly one element. Let x be the last element in Alice's array and y be the last element in Bob's array. Alice wants to maximize the absolute difference between x and y while Bob wants to minimize this value. Both players are playing optimally.

Find what will be the final value of the game.

Input

The first line contains a single integer n (1≤n≤1000) — the number of values in each array.

The second line contains nn space-separated integers a1,a2,…,an (1≤ai≤109) — the numbers in Alice's array.

The third line contains nn space-separated integers b1,b2,…,bn (1≤bi≤109) — the numbers in Bob's array.

Output

Print the absolute difference between xx and yy if both players are playing optimally.

Examples

Input

4
2 14 7 14
5 10 9 22

Output

4

Input

1
14
42

Output

28

Note

In the first example, the x=14x=14 and y=10y=10. Therefore, the difference between these two values is 44.

In the second example, the size of the arrays is already 11. Therefore, x=14x=14 and y=42y=42.

題目鏈接：https://vjudge.net/contest/345791#problem/G

試題解析 

我們先對a,b兩個數組求最小差值數組c，即ci=min(abs(ai-bj))。

我們知道存在bi使得ai與bi的差值最小，所以每當Alice去掉一個ai時，Bob就去掉與之對應的bi，為什么去掉bi？

當a數組與b數組取最小值是一一對應時，我們知道bi是使ai為最小的所以去掉ai后，bi不會使得a數組中其他數的最小值變得更小，所以Bob此時去掉相應的bi。Alice丟棄一個數時一定是將使得ci值最小的數去掉，到最后我們就會發現我們得到的值就是ci數組的中的最大值。

當a數組與b數組取最小值不是一一對應時，Alice還是會去掉使得ci值最小的那一個ai，當存在aj==ai時就說明會有一個b不會使得任何一個a最小，此時Bob就會去掉這個數b，到最后得到的值還是ci數組中的最大值。

所以我們只需要將ci數組求出來再找到最小值就好了。

 

#include<bits/stdc++.h>
using namespace std;
#define inf 0x3f3f3f3f
typedef long long ll;
const int mx=2e5+10;
int s1[mx],s2[mx];
int n;
 
int main()
{
	int ans=-1;
	scanf("%d",&n);
	for(int i=1;i<=n;i++)
	scanf("%d",&s1[i]);
	for(int i=1;i<=n;i++)
	scanf("%d",&s2[i]);
	for(int i=1;i<=n;i++)
	{
		int m=inf;
		for(int j=1;j<=n;j++)
		{
			m=min(m,abs(s1[i]-s2[j]));
		}
		ans=max(ans,m);
	}
	printf("%d\n",ans);
	return 0;
}