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C
Koxia and Number Theory
Codeforces
1700
原题
← BD →
翻译中…

Joi has an array $a$ of $n$ positive integers. Koxia wants you to determine whether there exists a positive integer $x \gt 0$ such that $\gcd(a_i+x,a_j+x)=1$ for all $1 \leq i \lt j \leq n$.

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Here $\gcd(y, z)$ denotes the greatest common divisor (GCD) of integers $y$ and $z$.

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Input

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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.

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The first line of each test case contains an integer $n$ ($2 \leq n \leq 100$) — the size of the array.

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The second line of each test case contains $n$ integers $a_1, a_2, \dots, a_n$ ($1 \leq a_i \leq {10}^{18}$).

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It is guaranteed that the sum of $n$ over all test cases does not exceed $1000$.

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Output

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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 $1 \leq i \lt j \leq n$, and "NO" (without quotes) otherwise.

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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.

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Note

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In the first test case, we can set $x = 4$. This is valid because:

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  • 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 $j=3$, $\gcd(a_i+x,a_j+x)=\gcd(7+4,10+4)=\gcd(11,14)=1$.
翻译中…

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.

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样例

Input #1
2
3
5 7 10
3
3 3 4
Output #1
YES
NO
InputCopy OutputCopy
题目信息
题号
1770C
时间限制
1 s
内存限制
256 MB
标签
暴力
中国剩余定理
数学
数论

题面与样例版权归 Codeforces 及出题人所有;中文翻译由 AI 生成(CI 管线),如有出入请以原文为准。