The Meeting Place Cannot Be Changed CodeForces - 782B （二分時間）

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.

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.

第一次作者種的二分，很是無奈，還
好已經做過了. 二分時間是比較簡單的一種方法，看是不是每個人走的區間是不是都能相交。

上代碼：

#include<cstdio>
#include<cstring>
#include<iostream>
#include<cmath>
#include<algorithm>
using namespace std;
int n;
double l;
double r;
struct node
{
    double x;
    double v;
};
node op[60001];
bool cmp(node a,node b)
{
    return a.x < b.x;
}
double x[60001];
double v[60001];
bool ok(double mid)
{
    double xx = x[1] - mid*v[1];
    double yy = x[1] + mid*v[1];
    for(int i=2; i<=n; i++)
    {
        double tx = x[i] - mid*v[i];
        double ty = x[i] + mid*v[i];
        if(tx <= xx && ty >= yy){;}
        else if(tx >= xx && ty >=yy && tx<= yy){xx=tx;}
        else if(tx <= xx && ty <= yy){yy=ty;}
        else if(tx >= xx&&ty<=yy){xx=tx;yy=ty;}
        else return false;
    }
    return true;
}
int main()
{
   while(scanf("%d", &n) != EOF)
   {
       double l;
       double r;
       for(int i=1; i<=n; i++)
       {
           scanf("%lf", &op[i].x);
       }
       for(int i=1; i<=n; i++)
       {
           scanf("%lf", &op[i].v);
       }
       sort(op+1, op+1+n,cmp);
       for(int i=1; i<=n; i++)
       {
           x[i] = op[i].x;
           v[i] = op[i].v;
       }
       l = 1e10;
       r = 0;
       double ans = 0;
       while(fabs(l-r) > 1e-6)
       {
          double mid = (l+r)/2.0;
          if(ok(mid))
            {l = mid;ans = mid;}
          else r=mid;
       }
       printf("%.7lf\n",ans);
   }
   return 0;
}

水波.