Bash’s Big Day
Time Limit: 2 Sec Memory Limit: 128 MB
Bash has set out on a journey to become the greatest Pokemon master. To get his first Pokemon, he went to Professor Zulu’s Lab. Since Bash is Professor Zulu’s favourite student, Zulu allows him to take as many Pokemon from his lab as he pleases.

But Zulu warns him that a group of k>1 Pokemon with strengths {s1,s2,s3,…,sk} tend to fight among each other if gcd(s1, s2, s3, …, sk) = 1 (see notes for gcd definition).

Bash, being smart, does not want his Pokemon to fight among each other. However, he also wants to maximize the number of Pokemon he takes from the lab. Can you help Bash find out the maximum number of Pokemon he can take?

Note: A Pokemon cannot fight with itself.

传送门:CF757B

Input

The input consists of two lines. The first line contains an integer n (1≤n≤105), the number of Pokemon in the lab. The next line contains n space separated integers, where the i-th of them denotes si (1≤si≤105), the strength of the i-th Pokemon.

Output

Print single integer — the maximum number of Pokemons Bash can take.

Sample Input

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2 3 4

Sample Output

1
2

题解

暴力把每个数的倍数算一下= =
## AC code:(不包含输入类)
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import java.io.*;  
import java.util.*;
public class Main {
static boolean isprime[]=new boolean [100005];
public static void main(String[] args) {
Scanner sc= new Scanner(System.in);
while(sc.hasNext()){
int n=sc.nextInt();
int[]shu=new int[100005];
for(int i=1;i<=n;i++) {
shu[sc.nextInt()]++;
}
int max=1;
for(int i=2;i<100005;i++){
int cur=0;
for(int j=i;j<100005;j=j+i){ //暴力求每个数和它的倍数
cur=cur+shu[j];
}
max=Math.max(max, cur);
}
System.out.println(max);
}

}
public static long gcd(long a,long b){
return a == 0 ? b : gcd(b % a, a);
}
}