Problem: While Grisha was celebrating New Year with Ded Moroz, Misha gifted Sasha a small rectangular pond of size n × m, divided into cells of size 1 × 1, inhabited by tiny evil fishes (no more than one fish per cell, otherwise they'll strife!).

The gift bundle also includes a square scoop of size r × r, designed for fishing. If the lower-left corner of the scoop-net is located at cell (x, y), all fishes inside the square (x, y)...(x + r - 1, y + r - 1) get caught. Note that the scoop-net should lie completely inside the pond when used.

Unfortunately, Sasha is not that skilled in fishing and hence throws the scoop randomly. In order to not frustrate Sasha, Misha decided to release k fishes into the empty pond in such a way that the expected value of the number of caught fishes is as high as possible. Help Misha! In other words, put k fishes in the pond into distinct cells in such a way that when the scoop-net is placed into a random position among (n - r + 1)·(m - r + 1) possible positions, the average number of caught fishes is as high as possible.

Input

The only line contains four integers n, m, r, k (1 ≤ n, m ≤ 105, 1 ≤ r ≤ min(n, m), 1 ≤ k ≤ min(n·m, 105)).

Output

Print a single number — the maximum possible expected number of caught fishes.

You answer is considered correct, is its absolute or relative error does not exceed 10 - 9. Namely, let your answer be a, and the jury's answer be b. Your answer is considered correct, if <image>.

Examples

Input

3 3 2 3


Output

2.0000000000


Input

12 17 9 40


Output

32.8333333333

Note

In the first example you can put the fishes in cells (2, 1), (2, 2), (2, 3). In this case, for any of four possible positions of the scoop-net (highlighted with light green), the number of fishes inside is equal to two, and so is the expected value.

<image>

Solution (in JAVA):

Solution:
import java.io.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.PriorityQueue;
import java.util.Scanner;
import java.util.Comparator;
import java.util.TreeSet;

/**
 * Built using CHelper plug-in
 * Actual solution is at the top
 */
public class Main {
	public static void main(String[] args) {
		InputStream inputStream = System.in;
		OutputStream outputStream = System.out;
		Scanner in = new Scanner(inputStream);
		PrintWriter out = new PrintWriter(outputStream);
		TaskD solver = new TaskD();
		solver.solve(1, in, out);
		out.close();
	}

	static class TaskD {
		private int n;
		private int m;
		private int r;
		private int k;

		public int e(int i, int j) {
			return (Math.min(n, i + r - 1) - Math.max(r, i) + 1) * (Math.min(m, j + r - 1) - Math.max(r, j) + 1);
		}

		public void solve(int testNumber, Scanner in, PrintWriter out) {
			//Since we don't need to get all the fish but the best ones, and k is fairly small, let's just generate the best fishes.
			n = in.nextInt();
			m = in.nextInt();
			r = in.nextInt();
			k = in.nextInt();
			long ans = 0;
			PriorityQueue<Point> pq = new PriorityQueue<>(new Comparator<Point>() {

				public int compare(Point o1, Point o2) {
					return e(o2.i, o2.j) - e(o1.i, o1.j);
				}
			});
			TreeSet<Point> visited = new TreeSet<>();
			pq.add(new Point((n + 1) / 2, (m + 1) / 2));
			while (k > 0) {
				Point p = pq.poll();
				if (visited.contains(p)) continue;
				ans += e(p.i, p.j);
				k--;
				visited.add(p);
				Point p1 = new Point(p.i - 1, p.j), p2 = new Point(p.i + 1, p.j), p3 = new Point(p.i, p.j - 1), p4 = new Point(p.i, p.j + 1);
				if (p.i > 1) pq.add(p1);
				if (p.i < n) pq.add(p2);
				if (p.j > 1) pq.add(p3);
				if (p.j < m) pq.add(p4);
			}
			out.println(ans / (double) ((long) (n - r + 1) * (m - r + 1)));
		}

		public class Point implements Comparable<Point> {
			int i;
			int j;

			public Point(int i, int j) {
				this.i = i;
				this.j = j;
			}


			public int compareTo(Point o) {
				return (i - o.i) * m + (j - o.j);
			}

		}

	}
}