CF782B The Meeting Place Cannot Be Changed

题意：

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

思路：

二分，注意控制精度。

实现：

 #include <cstdio>
 #include <iostream>
 #include <algorithm>
 #include <iomanip>
 using namespace std;
 struct node
 {
     double pos;
     double speed;
 };
 node a[];
 int n;
 double minn, maxn;
 bool check(double t)
 {
     minn = a[].pos - t * a[].speed;
     maxn = a[].pos + t * a[].speed;
     for (int i = ; i < n; i++)
     {
         double tmp_l = a[i].pos - t * a[i].speed;
         double tmp_r = a[i].pos + t * a[i].speed;
         minn = max(minn, tmp_l);
         maxn = min(maxn, tmp_r);
     }
     return maxn - minn >= 1e-;
 }

 double solve()
 {
     double l = 0.0, r = , res = ;
     for (int i = ; i < ; i++)
     {
         double mid = (l + r) / 2.0;
         if (check(mid))
         {
             r = mid;
             res = mid;
         }
         else
         {
             l = mid;
         }
     }
     return res;
 }

 int main()
 {
     cin >> n;
     for (int i = ; i < n; i++)
     {
         cin >> a[i].pos;
     }
     for (int i = ; i < n; i++)
     {
         cin >> a[i].speed;
     }
     cout << setprecision() << solve() << endl;
     return ;
 }