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
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.StringTokenizer;

public class Main {

	static BufferedReader in;
	static PrintWriter out;
	static StringTokenizer tok;

	void solve() throws IOException {
		solve(nl(), nl(), nl());
	}

	private void solve(long n, long m, long d) {

		if (2 * d >= n || d == 0)
			out.printf("%.10f\n", (double) (m - 1) / n);
		else
			out.printf("%.10f\n", (double) (2 * d + (n - 2 * d) * 2) * (m - 1) / n / n);

	}

	String ns() throws IOException {
		while (!tok.hasMoreTokens()) {
			tok = new StringTokenizer(in.readLine(), " ");
		}
		return tok.nextToken();
	}

	int ni() throws IOException {
		return Integer.parseInt(ns());
	}

	long nl() throws IOException {
		return Long.parseLong(ns());
	}

	double nd() throws IOException {
		return Double.parseDouble(ns());
	}

	String[] nsa(int n) throws IOException {
		String[] res = new String[n];
		for (int i = 0; i < n; i++) {
			res[i] = ns();
		}
		return res;
	}

	int[] nia(int n) throws IOException {
		int[] res = new int[n];
		for (int i = 0; i < n; i++) {
			res[i] = ni();
		}
		return res;
	}

	long[] nla(int n) throws IOException {
		long[] res = new long[n];
		for (int i = 0; i < n; i++) {
			res[i] = nl();
		}
		return res;
	}

	public static void main(String[] args) throws IOException {
		in = new BufferedReader(new InputStreamReader(System.in));
		out = new PrintWriter(System.out);
		tok = new StringTokenizer("");
		Main main = new Main();
		main.solve();
		out.close();
	}
}