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 <image>.

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.
import java.util.*;
import java.math.*;
import java.io.*;

public class solution {
    public static void merge(int arr[], int l, int m, int r) {
        // Find sizes of two subarrays to be merged
        int n1 = m - l + 1;
        int n2 = r - m;

        /* Create temp arrays */
        int L[] = new int[n1];
        int R[] = new int[n2];

        /*Copy data to temp arrays*/
        for (int i = 0; i < n1; ++i)
            L[i] = arr[l + i];
        for (int j = 0; j < n2; ++j)
            R[j] = arr[m + 1 + j];
        //        /* Merge the temp arrays */

        // Initial indexes of first and second subarrays
        int i = 0, j = 0;

        // Initial index of merged subarry array
        int k = l;
        while (i < n1 && j < n2) {
            if (L[i] <= R[j]) {
                arr[k] = L[i];
                i++;
            } else {
                arr[k] = R[j];
                j++;
            }
            k++;
        }

        /* Copy remaining elements of L[] if any */
        while (i < n1) {
            arr[k] = L[i];
            i++;
            k++;
        }

        /* Copy remaining elements of R[] if any */
        while (j < n2) {
            arr[k] = R[j];
            j++;
            k++;
        }
    }

    // Main function that sorts arr[l..r] using
    // merge()
    public static void sort(int arr[], int l, int r) {
        if (l < r) {
            // Find the middle point
            int m = (l + r) / 2;

            // Sort first and second halves
            sort(arr, l, m);
            sort(arr, m + 1, r);
            // Merge the sorted halves
            merge(arr, l, m, r);
        }
    }

    static class Pair {
        public int x1;
        public int x2;

        Pair(int x, int y) {
            this.x1 = x;
            this.x2 = y;
        }

        boolean comp(Pair A, Pair B) {
            return (A.x1 < B.x2);
        }
    }
    public  static double distance(double x,double x1,double y,double y1){
        double xx = Math.pow(x-x1,2);
        double yy = Math.pow(y-y1,2);
        double distance = Math.sqrt(xx+yy);
        return distance;
    }

    public static void main(String[] args) throws IOException {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        StringTokenizer st = new StringTokenizer(br.readLine());
        double x = Double.parseDouble(st.nextToken());
        double y = Double.parseDouble(st.nextToken());
        double min = Double.MAX_VALUE;
        int n = Integer.parseInt(br.readLine());
        for(int i=0;i<n;i++){
            st = new StringTokenizer(br.readLine());
            double x1 = Double.parseDouble(st.nextToken());
            double y1 = Double.parseDouble(st.nextToken());
            double ti = Double.parseDouble(st.nextToken());
            double dis = distance(x,x1,y,y1);
            double time = dis/ti;
            min = Math.min(min,time);
        }
        System.out.printf("%.10f", min);
        System.out.println();
    }
}