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.

 #include<cstdio>

 struct type{

     int x;

     int v;

 }p[+];

 int n;

 bool check(double time)

 {

     //遍历每个人在time个单位时间后能走到的位置，不断更新重叠的区间[l,r]，只要到最后这个区间一人不为空，就return true

     double l=p[].x-time*p[].v;

     double r=p[].x+time*p[].v;

     for(int i=;i<=n;i++){

         if(p[i].x-time*p[i].v > l) l=p[i].x-time*p[i].v;

         if(p[i].x+time*p[i].v < r) r=p[i].x+time*p[i].v;

         if(l>r) return false;

     }

     return true;

 }

 int main()

 {

     scanf("%d",&n);

     for(int i=;i<=n;i++) scanf("%d",&p[i].x);

     for(int i=;i<=n;i++) scanf("%d",&p[i].v);

     double st=,ed=;

     while(ed-st>1e-){

         double mid=st+(ed-st)/;

         if(check(mid)) ed=mid;

         else st=mid;

     }

     printf("%.12lf\n",ed);

 }