题目

Description

Vasiliy lives at point (a, b) of the coordinate plane. He is hurrying up to work so he wants to get out of his house as soon as possible. New app suggested n available Beru-taxi nearby. The i-th taxi is located at point (xi, yi) and moves with a speed vi.

Consider that each of n drivers will move directly to Vasiliy and with a maximum possible speed. Compute the minimum time when Vasiliy will get in any of Beru-taxi cars.

Input

The first line of the input contains two integers a and b ( - 100 ≤ a, b ≤ 100) — coordinates of Vasiliy's home.

The second line contains a single integer n (1 ≤ n ≤ 1000) — the number of available Beru-taxi cars nearby.

The i-th of the following n lines contains three integers xi, yi and vi ( - 100 ≤ xi, yi ≤ 100, 1 ≤ vi ≤ 100) — the coordinates of the i-th car and its speed.

It's allowed that several cars are located at the same point. Also, cars may be located at exactly the same point where Vasiliy lives.

Output

Print a single real value — the minimum time Vasiliy needs to get in any of the Beru-taxi cars. You answer will be considered correct if its absolute or relative error does not exceed 10 - 6.

Namely: let's assume that your answer is a, and the answer of the jury is b. The checker program will consider your answer correct, if .

Sample Input

Input

0 0
2
2 0 1
0 2 2

Output

1.00000000000000000000

Input

1 3
3
3 3 2
-2 3 6
-2 7 10

Output

0.50000000000000000000

Hint

In the first sample, first taxi will get to Vasiliy in time 2, and second will do this in time 1, therefore 1 is the answer.

In the second sample, cars 2 and 3 will arrive simultaneously.

题解

使用两点间距离公式计算距离
除以速度获得时间
输出最小的时间
与正确答案误差应该尽可能小,因此应该尽可能多输出小数位数

按照样例输出 20 位小数

代码

/*
By:OhYee
Github:OhYee
Blog:http://www.oyohyee.com/
Email:oyohyee@oyohyee.com

かしこいかわいい？
エリーチカ！
要写出来Хорошо的代码哦~
*/
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <cmath>
#include <string>
#include <iostream>
#include <vector>
#include <list>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <functional>
#include <bitset>
#include <iomanip> 
using namespace std;

inline double dis(double x1,double y1,double x2,double y2) {
    return sqrt((x1 - x2)*(x1 - x2) + (y1 - y2)*(y1 - y2));
}

bool Do() {
    double x,y;
    if(!(cin >> x >> y))
        return false;

    int n;
    cin >> n;

    double Min = 9999999999;
    for(int i = 1;i <= n;i++) {
        double tx,ty,v;
        cin >> tx >> ty >> v;
        double distance = fabs(dis(x,y,tx,ty));
        Min = min(Min,distance / v);
    }

    cout << fixed << setprecision(20) << Min << endl;

    return true;
}

int main() {
    cin.tie(0);
    cin.sync_with_stdio(false);

    while(Do());
    return 0;
}