A - A Compatible Pair

Problem description

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

252

Input

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

Output

2

Note

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

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

解题思路：题目给的数据很小，暴力枚举即可。题目的意思就是Big Banban从小Tommy的灯笼中选择一盏小Tommy没有钻进的灯笼，和自己的一盏灯笼组合后亮度不是最亮（即乘积的值不是最大），但Big Banban会选择一个组合，其亮度是最亮的（不能选小Tommy钻进的那个灯笼）。做法：c[i]存放a数组中当前第i盏灯笼和b数组中每盏灯笼组合后的最大亮度（乘积值最大），然后将c数组排序，c[n-1]即为小Tommy不让Big Banban选择的那个最大亮度的组合，但Big Banban可以选择剩下的最大值c[n-2]，即为那对灯笼组合的最大亮度。

AC代码：

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 typedef long long LL;
 4 int main(){
 5     int n,m;LL a[55],b[55],c[55];
 6     cin>>n>>m;
 7     for(int i=0;i<n;++i)cin>>a[i];
 8     for(int i=0;i<m;++i)cin>>b[i];
 9     for(int i=0;i<n;++i){
10         c[i]=a[i]*b[0];
11         for(int j=1;j<m;++j)
12             c[i]=max(c[i],a[i]*b[j]);
13     }
14     sort(c,c+n);cout<<c[n-2]<<endl;
15     return 0;
16 }