Codecraft-17 and Codeforces Round #391 (Div. 1 + Div. 2, combined) B. Bash's Big Day （Hash+简单数论）

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 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

Copy

3
2 3 4

output

Copy

2

input

Copy

5
2 3 4 6 7

output

Copy

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.

思路：先说暴力枚举做法，从2枚举到n个数中最大数，然后不断更新答案

然后就是换个思路枚举，降低下时间复杂度，cf的题目基本有一大半都是原本暴力超时然后换种方式有技巧的枚举就过了，这道题呢，可以枚举因数，最后统计最大的因数，因为因数的个数就是可能的情况，取最大就是所求的答案~

#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define rg register ll
#define inf 2147483647
#define lb(x) (x&(-x))
ll sz[200005],n;
template <typename T> inline void read(T& x)
{
    x=0;char ch=getchar();ll f=1;
    while(!isdigit(ch)){if(ch=='-')f=-1;ch=getchar();}
    while(isdigit(ch)){x=(x<<3)+(x<<1)+(ch^48);ch=getchar();}x*=f;
}
ll a[100005];
map<ll,ll>p;
int main()
{
	cin>>n;
	for(rg i=1;i<=n;i++)cin>>a[i];
	for(rg i=1;i<=n;i++)
	{
		for(rg j=1;j*j<=a[i];j++)
		{
			if(a[i]%j==0)
			{
				p[j]++;
				ll k=a[i]/j;
				if(k!=j)p[k]++;
			}
		}
	}
	ll maxx=1;
	for(auto it:p)
	{
		if(it.first!=1)
		maxx=max(it.second,maxx);
	}
	cout<<maxx<<endl;
    return 0;
    
}