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[60005];
int n;
double minn, maxn;
bool check(double t)
{
    minn = a[0].pos - t * a[0].speed;
    maxn = a[0].pos + t * a[0].speed;
    for (int i = 1; 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-7;
}

double solve()
{
    double l = 0.0, r = 1000000005, res = 1000000005;
    for (int i = 0; i < 10000; 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 = 0; i < n; i++)
    {
        cin >> a[i].pos;
    }
    for (int i = 0; i < n; i++)
    {
        cin >> a[i].speed;
    }
    cout << setprecision(7) << solve() << endl;
    return 0;
}