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.

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.

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.

37 1 31 2 1output
2.000000000000input
45 10 3 22 3 2 4output
1.400000000000Note
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.
ps：二分法枚舉時間，第i人在time時間內可到達的區間范圍為（x[i]-v[i]*time,x[i]+v[i]*time），然后判斷所有人都能到達的范圍是否有交集，如果有交集，說明他們都能在time時間內集合代碼：
#include<stdio.h>#define maxn 60000+10#define min(a,b) (a<b?a:b)#define max(a,b) (a>b?a:b)int x[maxn],v[maxn];int n;int canmeet(double time){    double left,right,le,ri;    for(int i=1; i<=n; i++)    {        if(i==1)            left=x[i]-v[i]*time,right=x[i]+v[i]*time;        else        {            le=x[i]-v[i]*time,ri=x[i]+v[i]*time;            if(le>right||ri<left)                return 0;            left=max(left,le);//維護區間最大的左邊值            right=min(ri,right);//維護區間最小的右邊值        }    }    return 1;}int main(){    scanf("%d",&n);    for(int i=1; i<=n; i++) scanf("%d",x+i);    for(int i=1; i<=n; i++) scanf("%d",v+i);    double l=0,r=1e9;    while(r-l>1e-6)    {        double mid=(l+r)/2;        if(canmeet(mid))//能在mid時間內能集合            r=mid;        else            l=mid;    }    printf("%.12lf/n",r);    return 0;}