CodeForces 706A Beru

CodeForces 706A Beru"/>

CodeForces 706A Beru

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 (x_i, y_i) and moves with a speed v_i.

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 x_i, y_i and v_i ( - 100 ≤ x_i, y_i ≤ 100, 1 ≤ v_i ≤ 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.

/*题意：已知你的坐标和n个司机的坐标和速度,求上车的最短时间类型：水题分析：直接写就行了
*/
#include<cstdio>
#include<algorithm>
#include<queue>
#include<iostream>
#include<cmath>
#include<cstring>
using namespace std;
int main()
{int sx,sy,x,y,v,m;scanf("%d%d",&sx,&sy);double Min=0x7fffffff;scanf("%d",&m);while(m--){scanf("%d%d%d",&x,&y,&v);double res=sqrt((sx-x)*(sx-x)+(sy-y)*(sy-y));if(res*1.0/v<Min)Min=res*1.0/v;}printf("%.7f\n",Min);return 0;
}