Codeforces 757B. Bash's Big Day GCD

B. Bash's Big Day

time limit per test：2 seconds

memory limit per test：512 megabytes

input：standard input

output：

standard output

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 ifgcd(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.

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题目链接：http://codeforces.com/problemset/problem/757/B

题意：给出n个数，求一个最大的集合并且这个集合中的元素gcd的结果不等于1。

思路：ai sign[ai]++;ai%j==0 sign[j]++,sign[ai/j]++;求最大的sign。注意sign[1]=1。时间复杂度n*sqrt(n);

代码：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
using namespace std;
const int MAXN=1e5+100;
int num[MAXN],sign[MAXN];
int main()
{
    int n;
    scanf("%d",&n);
    memset(sign,0,sizeof(sign));
    for(int i=1; i<=n; i++)
    {
        scanf("%d",&num[i]);
        sign[num[i]]++;
        int tmp=sqrt(num[i]);
        for(int j=2; j<=tmp; j++)
        {
            if(num[i]%j==0)
            {
                sign[j]++;
                if(num[i]/j!=j) sign[num[i]/j]++;
            }
        }
    }
    sign[1]=1;
    int ans=1;
    for(int i=1; i<=100000; i++)
        ans=max(ans,sign[i]);
    cout<<ans<<endl;
    return 0;
}