链接：https://codeforces.com/problemset/problem/1133/D

You are given two arrays aa and bb, each contains nn integers.

You want to create a new array cc as follows: choose some real (i.e. not necessarily integer) number dd, and then for every i∈[1,n]i∈[1,n] let ci:=d⋅ai+bici:=d⋅ai+bi.

Your goal is to maximize the number of zeroes in array cc. What is the largest possible answer, if you choose dd optimally?

Input

The first line contains one integer nn (1≤n≤2⋅1051≤n≤2⋅105) — the number of elements in both arrays.

The second line contains nn integers a1a1, a2a2, ..., anan (−109≤ai≤109−109≤ai≤109).

The third line contains nn integers b1b1, b2b2, ..., bnbn (−109≤bi≤109−109≤bi≤109).

Output

Print one integer — the maximum number of zeroes in array cc, if you choose dd optimally.

Examples

input

Copy

5
1 2 3 4 5
2 4 7 11 3

output

Copy

2

input

Copy

3
13 37 39
1 2 3

output

Copy

2

input

Copy

4
0 0 0 0
1 2 3 4

output

Copy

0

input

Copy

3
1 2 -1
-6 -12 6

output

Copy

3

Note

In the first example, we may choose d=−2d=−2.

In the second example, we may choose d=−113d=−113.

In the third example, we cannot obtain any zero in array cc, no matter which dd we choose.

In the fourth example, we may choose d=6d=6.

代码：

#include<bits/stdc++.h>
using namespace std;
long long n,t,l,r,k,s,d,ans,max1=0,mod=1e9+7;
long long b[500005],a[500005],dp[500005];
map<long double ,long long>m;
long double c[500005];
int main()
{
	cin>>n;
	d=0;
	for(int i=1;i<=n;i++)
	{
		cin>>a[i];
	}
	for(int i=1;i<=n;i++)
	{
		cin>>b[i];
	}
	for(int i=1;i<=n;i++)
	{
		if(b[i]==0&&a[i]==0)
		{
			d++;
		}
		else if(b[i]==0)
		{
			m[0.0]++;
		}
		else if(a[i]!=0)
		{
			c[i]=b[i]*1.0/a[i];
			m[c[i]]++;
		}
		max1=max(max1,m[c[i]]);
	}
	cout<<max1+d;
	
}