C. Koxia and Number Theorytime limit per test1 secondmemory limit per test256 megabytesinputstandard inputoutputstandard outputJoi has an array a of n positive integers. Koxia wants you to determine whether there exists a positive integer x > 0 such that \gcd(a_i+x,a_j+x)=1 for all 1 \leq i < j \leq n.Here \gcd(y, z) denotes the greatest common divisor (GCD) of integers y and z.InputEach test consists of multiple test cases. The first line contains a single integer t (1 \le t \le 100) — the number of test cases. The description of test cases follows.The first line of each test case contains an integer n (2 \leq n \leq 100) — the size of the array.The second line of each test case contains n integers a_1, a_2, \dots, a_n (1 \leq a_i \leq {10}^{18}).It is guaranteed that the sum of n over all test cases does not exceed 1000.OutputFor each test case, output "YES" (without quotes) if there exists a positive integer x such that \gcd(a_i+x,a_j+x)=1 for all 1 \leq i < j \leq n, and "NO" (without quotes) otherwise.You can output the answer in any case (upper or lower). For example, the strings "yEs", "yes", "Yes", and "YES" will be recognized as positive responses.ExampleInput
235 7 1033 3 4Output
YES
NO
NoteIn the first test case, we can set x = 4. This is valid because:   When i=1 and j=2, \gcd(a_i+x,a_j+x)=\gcd(5+4,7+4)=\gcd(9,11)=1.  When i=1 and j=3, \gcd(a_i+x,a_j+x)=\gcd(5+4,10+4)=\gcd(9,14)=1.  When i=2 and j=3, \gcd(a_i+x,a_j+x)=\gcd(7+4,10+4)=\gcd(11,14)=1. In the second test case, any choice of x makes \gcd(a_1 + x, a_2 + x) = \gcd(3+x,3+x)=3+x. Therefore, no such x exists.
235 7 1033 3 4
YES NO 
['brute force', 'chinese remainder theorem', 'math', 'number theory', '*1700']