A13738. Gifts Fixing

You have $n$ gifts and you want to give all of them to children. Of course, you don't want to offend anyone, so all gifts should be equal between each other. The $i$ -th gift consists of $a_i$ candies and $b_i$ oranges.

During one move, you can choose some gift $1 \le i \le n$ and do one of the following operations:

- eat exactly one candy from this gift (decrease $a_i$ by one);
- eat exactly one orange from this gift (decrease $b_i$ by one);
- eat exactly one candy and exactly one orange from this gift (decrease both $a_i$ and $b_i$ by one).

Of course, you can not eat a candy or orange if it's not present in the gift (so neither $a_i$ nor $b_i$ can become less than zero).

As said above, all gifts should be equal. This means that after some sequence of moves the following two conditions should be satisfied: $a_1 = a_2 = \dots = a_n$ and $b_1 = b_2 = \dots = b_n$ (and $a_i$ equals $b_i$ is not necessary).

Your task is to find the minimum number of moves required to equalize all the given gifts.

You have to answer $t$ independent test cases.

The first line of the input contains one integer $t$ ( $1 \le t \le 1000$ ) — the number of test cases. Then $t$ test cases follow.

The first line of the test case contains one integer $n$ ( $1 \le n \le 50$ ) — the number of gifts. The second line of the test case contains $n$ integers $a_1, a_2, \dots, a_n$ ( $1 \le a_i \le 10^9$ ), where $a_i$ is the number of candies in the $i$ -th gift. The third line of the test case contains $n$ integers $b_1, b_2, \dots, b_n$ ( $1 \le b_i \le 10^9$ ), where $b_i$ is the number of oranges in the $i$ -th gift.

For each test case, print one integer: the minimum number of moves required to equalize all the given gifts.

In the first test case of the example, we can perform the following sequence of moves:

- choose the first gift and eat one orange from it, so $a = [3, 5, 6]$ and $b = [2, 2, 3]$ ;
- choose the second gift and eat one candy from it, so $a = [3, 4, 6]$ and $b = [2, 2, 3]$ ;
- choose the second gift and eat one candy from it, so $a = [3, 3, 6]$ and $b = [2, 2, 3]$ ;
- choose the third gift and eat one candy and one orange from it, so $a = [3, 3, 5]$ and $b = [2, 2, 2]$ ;
- choose the third gift and eat one candy from it, so $a = [3, 3, 4]$ and $b = [2, 2, 2]$ ;
- choose the third gift and eat one candy from it, so $a = [3, 3, 3]$ and $b = [2, 2, 2]$ .