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brute force

chinese remainder theorem

math

number theory

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C. Koxia and Number Theory

time limit per test

1 secondmemory limit per test

256 megabytesinput

standard inputoutput

standard 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 $$$z$$$.

Input

Each 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 $$$1000$$$.

Output

For 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.

Example

Input

235 7 1033 3 4

Output

YES NO

Note

In 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.

Codeforces (c) Copyright 2010-2023 Mike Mirzayanov

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