Min Max Swap Solution|Codeforces Round #768 (div 2)

                                                 A. Min Max Swap

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

You are given two arrays a$a$ and b$b$ of n$n$ positive integers each. You can apply the following operation to them any number of times:

• Select an index i$i$ (1≤i≤n$1\le i\le n$) and swap ai$a_i$ with bi$b_i$ (i. e. ai$a_i$ becomes bi$b_i$ and vice versa).

Find the minimum possible value of max(a1,a2,…,an)⋅max(b1,b2,…,bn)$max(a_1,a_2,\dots ,a_n)\cdot max(b_1,b_2,\dots ,b_n)$ you can get after applying such operation any number of times (possibly zero).

Input

The input consists of multiple test cases. The first line contains a single integer t$t$ (1≤t≤100$1\le t\le 100$) — the number of test cases. Description of the test cases follows.

The first line of each test case contains an integer n$n$ (1≤n≤100$1\le n\le 100$) — the length of the arrays.

The second line of each test case contains n$n$ integers a1,a2,…,an$a_1,a_2,\dots ,a_n$ (1≤ai≤10000$1\le a_i\le 10000$) where ai$a_i$ is the i$i$-th element of the array a$a$.

The third line of each test case contains n$n$ integers b1,b2,…,bn$b_1,b_2,\dots ,b_n$ (1≤bi≤10000$1\le b_i\le 10000$) where bi$b_i$ is the i$i$-th element of the array b$b$.

Output

For each test case, print a single integer, the minimum possible value of max(a1,a2,…,an)⋅max(b1,b2,…,bn)$max(a_1,a_2,\dots ,a_n)\cdot max(b_1,b_2,\dots ,b_n)$ you can get after applying such operation any number of times.

Note

In the first test, you can apply the operations at indices 2$2$ and 6$6$, then a=[1,4,6,5,1,5]$a=[1,4,6,5,1,5]$ and b=[3,2,3,2,2,2]$b=[3,2,3,2,2,2]$, max(1,4,6,5,1,5)⋅max(3,2,3,2,2,2)=6⋅3=18$max(1,4,6,5,1,5)\cdot max(3,2,3,2,2,2)=6\cdot 3=18$.

In the second test, no matter how you apply the operations, a=[3,3,3]$a=[3,3,3]$ and b=[3,3,3]$b=[3,3,3]$ will always hold, so the answer is max(3,3,3)⋅max(3,3,3)=3⋅3=9$max(3,3,3)\cdot max(3,3,3)=3\cdot 3=9$.

In the third test, you can apply the operation at index 1$1$, then a=[2,2]$a=[2,2]$, b=[1,1]$b=[1,1]$, so the answer is max(2,2)⋅max(1,1)=2⋅1=2$max(2,2)\cdot max(1,1)=2\cdot 1=2$.

Solution:

This question solution is updated within contest time. please click on follow button and subscribe our channel.if you follow then get notification of answer as soon as possible.