codeforces#403—B题（二分，三分）

B. The Meeting Place Cannot Be Changed

time limit per test

5 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

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 x_i meters and can move with any speed no greater than v_i 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, ..., x_n (1 ≤ x_i ≤ 10⁹) — the current coordinates of the friends, in meters.

The third line contains n integers v₁, v₂, ..., v_n (1 ≤ v_i ≤ 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.

Your answer will be considered correct, if its absolute or relative error isn't greater than 10^- 6. Formally, let your answer be a, while jury's answer be b. Your answer will be considered correct if holds.

Examples

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.

题意：在x轴上有很多人，这些人都有对应的最大移动速度，想要使所有人运动至同一点，求出最小运动时间。

/*
当然也可以三分，这里采用另一种二分时间的方法
耗费某一个时间时，求出最南边可以到达的点及最北端可以到达的点
当有交点时，说明都可以到达
*/
#include<bits/stdc++.h>
using namespace std;
const int MAXN=60000+100;
const double EPS=1e-6;
int x[MAXN],v[MAXN];
int main()
{
   // freopen("data.in","r",stdin);
    int n;
    cin>>n;
    int xss=0x3f3f3f3f,xnn=-1;
    for(int i=0;i<n;i++){
        cin>>x[i];
        xss=min(xss,x[i]);
        xnn=max(xnn,x[i]);
    }
    int vmin=0x3f3f3f3f,vmax=-1;
    for(int i=0;i<n;i++){
        cin>>v[i];
        vmin=min(vmin,v[i]);
        vmax=max(vmax,v[i]);
    }
    double l=0,r=(xnn-xss)*1.0/2/vmin+1;//初始化l与r缩小搜索范围，也可不初始化，直接给r一个大数即可
    double res=0;
    double mid;
    double tmp;
    while(fabs(r-l)>=EPS){
        mid=(l+r)/2.0;
        //cout<<fixed<<setprecision(6)<<mid<<endl;
        double xn=0x3f3f3f3f,xs=-1;
        for(int i=0;i<n;i++){
                xs=max(xs,x[i]*1.0-v[i]*mid);
                xn=min(xn,x[i]*1.0+v[i]*mid);
        }
        if(xn>xs){
           r=mid;
        }
        else{
            l=mid;
        }
    }
    cout<<fixed<<setprecision(12)<<mid<<endl;

}