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, ..., s_k} tend to fight among each other if gcd(s₁, s₂, s₃, ..., s_k) = 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 s_i (1 ≤ s_i ≤ 10⁵), the strength of the i-th Pokemon.

Output

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

Example

Input

3
2 3 4

Output

2

Input

5
2 3 4 6 7

Output3


题意：
给你一组数，求出里面最大公约数不为1的最多数量。

这道题刚开始最直观思路就是两个for循环，直接查找对于每一个数都求一下。当然这肯定超时了，
后来上网搜了一下，暴力枚举就OK了，思路都差不多，但是算法改进一下，对每个输入的数字求出因子，然后对因子的数量加1，这样只用一个for循环就OK了。

具体代码如下：

#include<iostream>
#include<algorithm>
#include<cstdio>
using namespace std;
const int INF = 100010;
int s[INF] = { 0 };
int main()
{
    int n;
    cin >> n;
    int m;
    for (int i = 0; i < n; i++)
    {
        scanf("%d", &m);
        int j = 1;
        for (j = 1; j * j < m; j++)//从1开始是对每个数都让自己这个数量从0变为1
        {
            if (m % j == 0)
            {
                s[j]++;
                s[m / j]++;//因子++
            }
        }
        if (j * j == m)
        {
            s[j]++;
        }
    }
    int ans = 1;
    for (int i = 2; i < INF; i++)
    {
        ans = max(ans, s[i]);
    }
    cout << ans << endl;
    return 0;
}