Codeforces Round #403 (based on Technocup 2017 Finals)

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 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.

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：给你n个人的坐标，速度，每个人运动的方向任意，请问让这些人相遇的最短时间是多少（在同一个位置相遇）

思路：三分位置。。。。首先，在坐标内任选一个位置，到这个位置所花时间最长的那个人是最后一个到达的，也就是这个人所花的时间是所有人到达这个位置所需要的时间。每个人到达各个位置与时间的函数是一条凹曲线，（如下图1），将所有人的曲线放在同一个图上，并且没个点取这个点的最大值，的到的曲线便是所要进行三分的曲线（如下图，加粗部分是三分曲线）

AC代码：

#include "iostream"
#include "string.h"
#include "stack"
#include "queue"
#include "string"
#include "vector"
#include "set"
#include "map"
#include "algorithm"
#include "stdio.h"
#include "math.h"
#define ll long long
#define mem(a) memset(a,0,sizeof(a))
using namespace std;
const double eps=1e-6;
struct Node{
    double x,v;
    bool friend operator< (Node a, Node b){
        return a.x<b.x;
    }
};
Node arr[200105];
int n;
double ans;
double fun(double x){
    double ans=0.0;
    for(int i=1; i<=n; ++i)
        if(ans<(fabs(arr[i].x-x)/arr[i].v))
            ans=(fabs(arr[i].x-x)/arr[i].v);
    return ans;
}
int main(){
    cin>>n;
    for(int i=1; i<=n; ++i) scanf("%lf",&arr[i].x);
    for(int i=1; i<=n; ++i) scanf("%lf",&arr[i].v);
    sort(arr+1,arr+n+1);
    double l=arr[1].x,r=arr[n].x;
    double mid,mmid,mv,mmv;
    while(l+eps<r){
        mid=(l+r)/2.0;
        mmid=(mid+r)/2.0;
        mv=fun(mid),mmv=fun(mmid);
        if(mmv>mv) r=mmid;
        else l=mid;
    }
    printf("%.12lf\n",fun(l));
    return 0;
}