CodeForces 706A Beru-taxi

A. Beru-taxi

time limit per test
1 second

memory limit per test
256 megabytes

input
standard input

output
standard output

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 

.

Examples

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

Note

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.
水题：
题意：给出Vasiliy家的坐标，和m个出租车的坐标以及速度，求出租车到Vasiliy家的最小时间

#include<stdio.h>
#include<math.h>
#include<string.h>
#include<algorithm>
#define INF 0xfffffff
using namespace std;
double dis(double x1,double y1,double x2,double y2)
{
return sqrt((x1-x2)*(x1-x2)+(y1-y2)*(y1-y2));
}
int main()
{
int n,i,j;
double x,y,a,b,v1,t,d;
while(scanf("%lf%lf",&x,&y)!=EOF)
{
double Min=INF;
scanf("%d",&n);
for(i=0;i<n;i++)
{
scanf("%lf%lf%lf",&a,&b,&v1);
d=dis(x,y,a,b);
t=d/v1;
if(t<Min)
Min=t;
}
printf("%lf\n",Min);
}
return 0;
}