Let us define the beauty of a sequence (a_1,... ,a_n) as the number of pairs of two adjacent elements in it whose absolute differences are d. For example, when d=1, the beauty of the sequence (3, 2, 3, 10, 9) is 3.

There are a total of n^m sequences of length m where each element is an integer between 1 and n (inclusive). Find the beauty of each of these n^m sequences, and print the average of those values.

Constraints

* 0 \leq d < n \leq 10^9
* 2 \leq m \leq 10^9
* All values in input are integers.

Input

Input is given from Standard Input in the following format:


n m d


Output

Print the average of the beauties of the sequences of length m where each element is an integer between 1 and n. The output will be judged correct if the absolute or relative error is at most 10^{-6}.

Examples

Input

2 3 1


Output

1.0000000000


Input

1000000000 180707 0


Output

0.0001807060
/*
* @Author Silviase(@silviasetitech)
* For ProCon
*/

import java.util.*;
import java.lang.*;
import java.math.*;

class Main{

        static int MOD = (int)1e9+7;

        // for dfs
        static int n;
        static int ansi;
        static int[] w;
        static int[] ww;
        static boolean[] isvisited;
        //
    public static void main(String[] args){
        Scanner sc = new Scanner(System.in);

        double n = sc.nextDouble();
        double m = sc.nextDouble();
        double d = sc.nextDouble();
        if (d == 0) {
            System.out.println( (m-1)/n);
        } else {
            System.out.println( 2*(m-1)*(n-d)/(n*n) );
        }

        sc.close();
    }

    private static long[][] CombinationLong_nCk(long n, long k) {
        n++; k++;
        long[][] ans = new long[(int)n][(int)k];
        for (int i = 0; i < n; i++) {
            ans[i][0] = 1;
        }
        for (int i = 0; i < n-1; i++) {
            for (int j = 0; j < k-1; j++) {
                if (i < j) {
                    ans[i][j] = 0;
                }else{
                    ans[i+1][j+1] = ans[i][j] + ans[i][j+1];
                }
            }
        }
        return ans;
    }

        public static int gcd_int(int a, int b) {
        
            if (a < b) {
                int tmpp = a;
                a = b;
                b = tmpp;
            }
            if (b == 0) {
                return a;
            }else{
                return gcd_int(b, a%b);
            }
        } 

        public static long gcd_long(long a, long b) {
        
            if (a < b) {
                long tmpp = a;
                a = b;
                b = tmpp;
            }
            if (b == 0) {
                return a;
            }else{
                return gcd_long(b, a%b);
            }
        }

        public static int lcm_int(int a, int b) {
            int gcd = gcd_int(a,b);
            return a/gcd*b;
        }

        public static long lcm_long(long a, long b) {
            long gcd = gcd_long(a,b);
            return a/gcd*b;
        }

        public static void dfs(int placenow) {
            // if went all -> success! 
            // if not went all -> fail...
            
            /*
            dfs    
            Go All Place that there is way to and not having been yet. 

             if island 1 is start point, dfs(1);
            if there is way to island 2 and island 3,
            - island 2 changes non-visited -> visited, and dfs(2).
            - island 3 changes non-visited -> visited, and dfs(3).
    
            */

            isvisited[placenow] = true;
            boolean success = true;
            for (int i = 0; i < n; i++) {
                if (isvisited[i] == false) { // not go all place
                    success = false;
                    break;
                }
            }
            if (success) {
                ansi++;
                isvisited[placenow] = false;
                return;
            }

            for (int i = 0; i < n; i++) {
                if (w[i] == placenow && isvisited[ww[i]] == false ) {
                    dfs(ww[i]);
                }else if(ww[i] == placenow && isvisited[w[i]] == false){
                    dfs(w[i]);
                }else{
                    continue;
                }
            }
            isvisited[placenow] = false;
            return;
        }

}

