Codeforces 757 B Bash's Big Day

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.

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.

Examples

input

3
2 3 4

output

2

input

5
2 3 4 6 7

output

3

Note

gcd (greatest common divisor) of positive integers set {a1, a2, ..., an} is
the maximum positive integer that divides all the integers {a1, a2, ..., an}.

In the first sample, we can take Pokemons with strengths {2, 4} since gcd(2, 4) = 2.

In the second sample, we can take Pokemons with strengths {2, 4, 6}, and there is no larger group with gcd ≠ 1.

题目大意：给你一串数，让你从中间挑出最多有几个数，他们的最大公因子数且不为1.

题目分析：直接枚举，不过貌似有个坑点，就是那个如果是给的数都是1的话，答案就是1.#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#include <queue>
#include <cmath>
using namespace std;
const int maxn = 100005;
int s[maxn];
int main(){
int n;
int k,m;
while((scanf("%d",&n))!=EOF){
memset(s,0,sizeof(s));
for(int i=1;i<=n;i++){
scanf("%d",&k);
int j=1;
for(;j*j<k;j++){
if(k % j == 0){
s[j]++;
s[k/j]++;
}
}

if(j*j == k)
s[j]++;

}
int ans=1;
for(int i=2; i<=maxn; i++)
ans=max(ans,s[i]);
printf("%d\n",ans);
}
return 0;
}