CF1631A.Min Max Swap

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题目描述

You are given two arrays aa and bb of nn positive integers each. You can apply the following operation to them any number of times:

  • Select an index ii ( 1in1\leq i\leq n ) and swap aia_i with bib_i (i. e. aia_i becomes bib_i and vice versa).

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

输入格式

The input consists of multiple test cases. The first line contains a single integer tt ( 1t1001 \leq t \leq 100 ) — the number of test cases. Description of the test cases follows.

The first line of each test case contains an integer nn ( 1n1001\le n\le 100 ) — the length of the arrays.

The second line of each test case contains nn integers a1,a2,,ana_1, a_2, \ldots, a_n ( 1ai100001 \le a_i \le 10\,000 ) where aia_i is the ii -th element of the array aa .

The third line of each test case contains nn integers b1,b2,,bnb_1, b_2, \ldots, b_n ( 1bi100001 \le b_i \le 10\,000 ) where bib_i is the ii -th element of the array bb .

输出格式

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, \ldots, a_n) \cdot \max(b_1, b_2, \ldots, b_n) you can get after applying such operation any number of times.

输入输出样例

  • 输入#1

    3
    6
    1 2 6 5 1 2
    3 4 3 2 2 5
    3
    3 3 3
    3 3 3
    2
    1 2
    2 1

    输出#1

    18
    9
    2

说明/提示

In the first test, you can apply the operations at indices 22 and 66 , 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)=63=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)=33=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 11 , then a=[2,2]a = [2, 2] , b=[1,1]b = [1, 1] , so the answer is max(2,2)max(1,1)=21=2\max(2, 2) \cdot \max(1, 1) = 2 \cdot 1 = 2 .

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