Codeforces 934.A A Compatible Pair

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, ..., an, and Banban's have brightness b1, b2, ..., bm 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, ..., an.

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

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.

题目大意：Tommy有n个数,Banban有m个数，Tommy会选择自己的一个数隐藏起来，Banban会选择自己的一个数和Tommy的一个没有隐藏的数相乘，Tommy想让结果最小，Banban想让结果最大，问最后的结果是什么.

分析：这题坑啊.以为只用判断两个端点的情况的，结果没想到有负数.

　　　这道题的话一个三重循环就能搞定.枚举隐藏哪一个数，将之后所得的乘积排个序，取第二大的就是答案.

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>

using namespace std;

typedef long long ll;

int n,m;
ll a[60],b[60],c[10010],tot,ans = 9223372036854775807;

int main()
{
    scanf("%d%d",&n,&m);
    for (int i = 1; i <= n; i++)
        cin >> a[i];
    for (int i = 1; i <= m; i++)
        cin >> b[i];
    for (int i = 1; i <= n; i++)
    {
        tot = 0;
        for (int j = 1; j <= n; j++)
        {
            if (j == i)
                continue;
            for (int k = 1; k <= m; k++)
                c[++tot] = a[j] * b[k];
        }
        sort(c + 1,c + tot + 1);
        ans = min(ans,c[tot]);
    }
    cout << ans << endl;

    return 0;
}