9660: ABC182 —— B - Almost GCD
[Creator : ]
Description
Given is an integer sequence $A$: $A_1, A_2, A_3, \dots, A_N$.
Let the GCD-ness of a positive integer $k$ be the number of elements among $A_1, A_2, A_3, \dots, A_N$ that are divisible by $k$.
Among the integers greater than or equal to $2$, find the integer with the greatest GCD-ness. If there are multiple such integers, you may print any of them.
Let the GCD-ness of a positive integer $k$ be the number of elements among $A_1, A_2, A_3, \dots, A_N$ that are divisible by $k$.
Among the integers greater than or equal to $2$, find the integer with the greatest GCD-ness. If there are multiple such integers, you may print any of them.
Input
Input is given from Standard Input in the following format:
```
$N$
$A_1 \hspace{7pt} A_2 \hspace{7pt} A_3 \hspace{5pt} \dots \hspace{5pt} A_N$
```
```
$N$
$A_1 \hspace{7pt} A_2 \hspace{7pt} A_3 \hspace{5pt} \dots \hspace{5pt} A_N$
```
Output
Print an integer with the greatest GCD-ness among the integers greater than or equal to $2$. If there are multiple such integers, any of them will be accepted.
Constraints
- $1 \le N \le 100$
- $2 \le A_i \le 1000$
- All values in input are integers.
- $2 \le A_i \le 1000$
- All values in input are integers.
Sample 1 Input
3
3 12 7
Sample 1 Output
3
Among 3, 12, and 7, two of them - 3 and 12 - are divisible by 3, so the GCD-ness of 3 is 2.
No integer greater than or equal to 2 has greater GCD-ness, so 3 is a correct answer.
No integer greater than or equal to 2 has greater GCD-ness, so 3 is a correct answer.
Sample 2 Input
5
8 9 18 90 72
Sample 2 Output
9
In this case, the GCD-ness of 9 is 4.
2 and 3 also have the GCD-ness of 4, so you may also print 2 or 3.
2 and 3 also have the GCD-ness of 4, so you may also print 2 or 3.
Sample 3 Input
5
1000 1000 1000 1000 1000
Sample 3 Output
1000