# The Meeting Place Cannot Be Changed （codeforces 782B）（二分法）

The main road in Bytecity is a straight line from south to north. Conveniently, there are coordinates measured in meters from the southernmost building in north direction.

At some points on the road there are n friends, and i-th of them is standing at the point xi meters and can move with any speed no greater than vi meters per second in any of the two directions along the road: south or north.

You are to compute the minimum time needed to gather all the n friends at some point on the road. Note that the point they meet at doesn’t need to have integer coordinate.

Input

The first line contains single integer n (2 ≤ n ≤ 60 000) — the number of friends.

The second line contains n integers x1, x2, …, xn (1 ≤ xi ≤ 109) — the current coordinates of the friends, in meters.

The third line contains n integers v1, v2, …, vn (1 ≤ vi ≤ 109) — the maximum speeds of the friends, in meters per second.

Output

Print the minimum time (in seconds) needed for all the n friends to meet at some point on the road.

Example

Input
3
7 1 3
1 2 1
Output
2.000000000000
Input
4
5 10 3 2
2 3 2 4
Output
1.400000000000

Note

In the first sample, all friends can gather at the point 5 within 2 seconds. In order to achieve this, the first friend should go south all the time at his maximum speed, while the second and the third friends should go north at their maximum speeds.

#include<iostream>
#include<cstdio>
double a[60001];
double v[60001];
int n;
using namespace std;
int judge(double x){//正义的审判
double left[n];
double right[n];

for(int i=0;i<n;i++){
left[i]=a[i]-x*v[i];
right[i]=a[i]+x*v[i];

}

double leftmax=left[0],rightmax=right[0];

for(int i=0;i<n;i++){//判断每个人是否都可以到达某一点（是否有公共区间）
leftmax=left[i]>leftmax?left[i]:leftmax;
rightmax=right[i]>rightmax?rightmax:right[i];

}

if(leftmax<rightmax){

return 1;
}else{
return 0;
}

}
int main(){
ios::sync_with_stdio(false);
cin>>n;
double min=10e9;
double max=a[0];
for(int i=0;i<n;i++){
cin>>a[i];
min=min>a[i]?a[i]:min;
max=max>a[i]?max:a[i];
}
double lowwer_speed=10e9;
for(int i=0;i<n;i++){
cin>>v[i];
lowwer_speed=lowwer_speed>v[i]?v[i]:lowwer_speed;
}

double l=0;
double r=(max-min)/lowwer_speed;
double mid=(r-l)/2;

while(r-l>1e-6){
mid=(r+l)/2;
if(judge(mid)){

r=mid;
}else{
l=mid;
}
}
printf("%.10llf\n",(r+l)/2);//注意小数点后的位数
return 0;
}