This is the easy version of the problem. The difference between the versions is that in this version, (b_i = 1) for all (i) ((1 \le i \le n)). You can hack only if you solved all versions of this problem. You find yourself with two arrays of positive integers (a) and (b), both of length (n). You will perform the following operation any number of times (possibly none): select an integer (i) ((1 \le i \le n)) and increase (a_i) by (1). This has a cost of (b_i). Determine the minimum total cost to make it so that there exists two integers (i, j) where (1 \le i < j \le n) and (\gcd(a_i, a_j))(^{\text{∗}})(> 1). (^{\text{∗}})(\gcd(x, y)) denotes the greatest common divisor (GCD) of integers (x) and (y). Each test contains multiple test cases. The first line contains the number of test cases (t) ((1 \le t \le 10^4)). The description of the test cases follows. The first line of each test case contains an integer (n) ((2 \le n \le 2 \cdot 10^5)) — the length of the array (a). The second line of each test case contains (n) integers (a_1,a_2,\ldots,a_n) ((1 \le a_i \le 2 \cdot 10^5)). The third line of each test case contains (n) integers (b_1,b_2,\ldots,b_n) ((\color{red}{b_i = 1})). The sum of (n) across all test cases does not exceed (2 \cdot 10^5). For each testcase, output the minimum cost. In the first test case, we can do the following: (\color{red}1, 1 \xrightarrow{x = 1} 2, \color{red}1 \xrightarrow{x = 2} 2, 2). Now (\gcd(a_1, a_2) = \gcd(2, 2) = 2) and so (\gcd(a_1, a_2) > 1). It can be proven that this is the minimum cost required. In the second test case, it is already true that (\gcd(a_1, a_2) = 4) and so (\gcd(a_1, a_2) > 1). So no operations are required.