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.
比較通俗的二分方法，但是因?yàn)榕袛嗄睦镂矣袀€(gè)位置放反了，導(dǎo)致測(cè)試數(shù)據(jù)出錯(cuò)，然后zz的以為是精度的問題，開大精度導(dǎo)致system testing的時(shí)候TLE。zz啊！
#include <map>#include <set>#include <cstdio>#include <cstring>#include <algorithm>#include <queue>#include <iostream>#include <stack>#include <cmath>#include <string>#include <vector>#include <cstdlib>//#include <bits/stdc++.h>//#define LOACL#define space " "using namespace std;//typedef long long LL;//typedef __int64 Int;typedef pair<int, int> PAI;const int INF = 0x3f3f3f3f;const double ESP = 1e-6;const double PI = acos(-1.0);const int MOD = 1e9 + 7;const int MAXN = 60000 + 10;double v[MAXN], ar[MAXN];int N;bool judge(long double t) {    long double r = ar[0] + v[0]*t;    long double l = ar[0] - v[0]*t;    for (int i = 1; i < N; i++) {        long double tr = ar[i] + v[i]*t;        long double tl = ar[i] - v[i]*t;        if (tl <= l && tr >= r) continue;        else if (tl >= l && tr <= r) {l = tl; r = tr;}        else if (tl <= l && l <= tr) {l = l; r = tr;}        else if (l <= tl && r >= tl) {l = tl; r = r;}        else return false;    }    return true;}int main() {    while (scanf("%d", &N) != EOF) {        for (int i = 0; i < N; i++) scanf("%d", &ar[i]);        for (int i = 0; i < N; i++) scanf("%d", &v[i]);        double ub = 1e9, lb = 0;        double ans = 0;        while (abs(ub - lb) > ESP) {            long double mid = (lb + ub)/2;            if (judge(mid)) {                ans = mid; ub = mid;            }            else lb = mid;        }        printf("%.7lf/n", ans);    }    return 0;}