cf462 A Compatible Pair

最新推荐文章于 2021-03-18 20:55:30 发布

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本文介绍了一个关于寻找两个集合中能组成最小亮度乘积配对的问题。通过隐藏其中一个集合的一个元素来达到目标，同时考虑了两个参与者的对立策略。

Nian is a monster which lives deep in the oceans. Once a year, it shows up on the land, devouring livestock and even people. In order to keep the monster away, people fill their villages with red colour, light, and cracking noise, all of which frighten the monster out of coming.

Little Tommy has n lanterns and Big Banban has m lanterns. Tommy's lanterns have brightness a₁, a₂, ..., a_n, and Banban's have brightness b₁, b₂, ..., b_m respectively.

Tommy intends to hide one of his lanterns, then Banban picks one of Tommy's non-hidden lanterns and one of his own lanterns to form a pair. The pair's brightness will be the product of the brightness of two lanterns.

Tommy wants to make the product as small as possible, while Banban tries to make it as large as possible.

You are asked to find the brightness of the chosen pair if both of them choose optimally.

Input

The first line contains two space-separated integers n and m (2 ≤ n, m ≤ 50).

The second line contains n space-separated integers a₁, a₂, ..., a_n.

The third line contains m space-separated integers b₁, b₂, ..., b_m.

All the integers range from - 10⁹ to 10⁹.

Output

Print a single integer — the brightness of the chosen pair.

Examples

Input

2 2
20 18
2 14

Output

Input

5 3
-1 0 1 2 3
-1 0 1

Output

Note

In the first example, Tommy will hide 20 and Banban will choose 18 from Tommy and 14 from himself.

In the second example, Tommy will hide 3 and Banban will choose 2 from Tommy and 1 from himself.

#include<iostream>
#include<algorithm>
using namespace std;
typedef long long ll;
#define f(i,l,r) for(int i=l;i<=r;i++)
#define g(i,l,r) for(int i=l;i>=r;i--)
const int maxn 	= 2505;
ll a[maxn],b[maxn],c[maxn];
ll n,m ; 
 
 int main()
 {
 	cin>>n>>m;
 	f(i,1,n)cin>>a[i];
 	f(i,1,m)cin>>b[i];
 	ll ans = -1000001;
 	ll t ;
	f(i,1,n) f(j,1,m)
	{
		if(ans<a[i]*b[j])
		{
			ans=a[i]*b[j];
			t= i;
		}
	}
	ans = -1000001;
	f(i,1,n)
	{
		if(i==t)continue;

		f(j,1,m)
		{
			if(ans<a[i]*b[j])
			{
				ans=a[i]*b[j]; 	 
			}
		}
	} 

	cout<<ans<<endl;

 }