Codeforces Round #403 (Div. 2, based on Technocup 2017 Finals) B. The Meeting Place Cannot Be Change

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, andi-th of them is standing at the point x_i meters and can move with any speed no greater thanv_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 than10^- 6. Formally, let your answer bea, while jury's answer be b. Your answer will be considered correct if holds.

Examples

Input

37 1 31 2 1

Output

2.000000000000

Input

45 10 3 22 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.

给定人的位置和移动速度,求最快移动到同一点的时间.在一条直线上..

我直接排序搞过了...因为出题人没构造出可以卡掉我的数据.数据比较难造就是了.

直接找所有人里面相遇的最大时间,我的阀值取了200..

正解是二分时间.假设所有人向左走,取最大坐标.假设所有人向右走.取最小坐标,然后向左的坐标大于向右的说明时间可行..

不是正解的诡异AC..

#include <cstdio>#include <cstring>#include <algorithm>#include <cmath>#include <iostream>using namespace std;#define met(a,b) memset(a,b,sizeof(a))#define inf 0x3f3f3f3fconst int maxn =1e3+10;typedef long long  ll;struct node {    double pos,v;} f[60005];bool cmp(node a,node b) {    if(a.v!=b.v)        return a.v<b.v;    else        return a.pos<b.pos;}bool cmp1(node a,node b) {    if(a.pos!=b.pos)        return a.pos<b.pos;    else        return a.v<b.v;}int main() {    int n;    while(cin>>n) {        for(int i=1; i<=n; i++) {            cin>>f[i].pos;        }        for(int i=1; i<=n; i++) {            cin>>f[i].v;        }        sort(f+1,f+1+n,cmp);        double time=0;        for(int j=1; j<=min(200,n); j++)            for(int i=j+1; i<=n; i++)                time=max(time,fabs(f[j].pos-f[i].pos)/fabs(f[j].v+f[i].v));        sort(f+1,f+1+n,cmp1);        for(int j=1; j<=min(200,n); j++)            for(int i=j+1; i<=n; i++)                time=max(time,fabs(f[j].pos-f[i].pos)/fabs(f[j].v+f[i].v));        printf("%.10lf\n",time);    }    return 0;}