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 ≤ 10^5, 1 ≤ r ≤ min(n, m), 1 ≤ k ≤ min(n·m, 10^5)).


-----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 $\frac{|a - b|}{\operatorname{max}(1,|b|)} \leq 10^{-9}$.


-----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]

Difficulty: HARD
Tags: Graph algorithms

Probability

Shortest paths

Greedy algorithms

Data structures
Skill types: Data structures

Greedy algorithms

Starter code:

Solutions:
import heapq
from heapq import heappush as push
from heapq import heappop as pop
base = 100009
Q = []
used = {}
ans = 0
dx = [0, 0, -1, 1]
dy = [-1, 1, 0, 0]

def f(x, y):
	(x1, x2) = (min(x, r - 1), max(0, x - n + r))
	(y1, y2) = (min(y, r - 1), max(0, y - m + r))
	return -(x1 - x2 + 1) * (y1 - y2 + 1)

def cal(i, j):
	return i * base + j

def get(ind):
	return (ind // base, ind % base)

def is_ok(x, y, x_, y_):
	if x + x_ >= 0 and x + x_ < n and (y + y_ >= 0) and (y + y_ < m):
		return True
	return False
(n, m, r, k) = map(int, input().split())
(x0, y0) = (n // 2, m // 2)
Q = [(f(x0, y0), cal(x0, y0))]
while k > 0:
	(val, ind) = pop(Q)
	if ind in used:
		continue
	(x, y) = get(ind)
	used[ind] = True
	val = -val
	ans += val
	for (x_, y_) in zip(dx, dy):
		if is_ok(x, y, x_, y_):
			push(Q, (f(x + x_, y + y_), cal(x + x_, y + y_)))
	k -= 1
print(ans / ((n - r + 1) * (m - r + 1)))


import queue
q = queue.PriorityQueue()
(n, m, r, k) = map(int, input().split())
a = min(r, n - r + 1)
b = min(r, m - r + 1)
u = n - 2 * a + 2
v = m - 2 * b + 2
for i in range(1, a + 1):
	q.put((-b * i, i))
t = 0
while k:
	(s, i) = q.get()
	q.put((s + i, i))
	d = min((u if i == a else 2) * (v if s == -b * i else 2), k)
	t -= s * d
	k -= d
print(t / (n - r + 1) / (m - r + 1))


import heapq as hq
from queue import PriorityQueue
import math
(n, m, r, k) = input().split()
N = int(n)
M = int(m)
R = int(r)
K = int(k)
q = PriorityQueue()
for i in range(1, math.floor((N + 1) / 2) + 1):
	maxi = min(min(i, N - i + 1), min(R, N - R + 1)) * min(min(R, M - R + 1), math.ceil(M / 2))
	num = M - (2 * min(min(R, M - R + 1), math.ceil(M / 2)) - 2)
	mult = 2
	if i > math.floor(N / 2):
		mult = 1
	q.put((-maxi, num * mult, i))
ans = 0
while K > 0:
	pop = q.get()
	a = -1 * pop[0]
	b = pop[1]
	c = pop[2]
	d = min(min(c, N - c + 1), min(R, N - R + 1))
	if d != a:
		mult = 2
		if c > N / 2:
			mult = 1
		q.put((-(a - d), 2 * mult, c))
	ans += a * min(b, K)
	K -= b
tot = (N - R + 1) * (M - R + 1)
print(str(ans / tot))


class TupleHeap(object):

	def __init__(self):
		self.heap = [0]

	def mult(self, i):
		return self.heap[i][0] * self.heap[i][1]

	def __str__(self):
		return ' '.join(map(str, self.heap))

	def add(self, var):
		self.heap.append(var)
		if len(self.heap) > 2:
			el = len(self.heap) - 1
			par = el // 2
			while el != 1 and self.mult(el) > self.mult(par):
				(self.heap[el], self.heap[par]) = (self.heap[par], self.heap[el])
				el = par
				par = el // 2

	def pop(self):
		if len(self.heap) == 1:
			return None
		ans = self.heap[1]
		if len(self.heap) == 2:
			self.heap.pop()
			return ans
		self.heap[1] = self.heap.pop()
		el = 1
		var_el = self.mult(el)
		child1 = el * 2
		if child1 < len(self.heap):
			var_child1 = self.mult(child1)
		else:
			var_child1 = -1
		child2 = el * 2 + 1
		if child2 < len(self.heap):
			var_child2 = self.mult(child2)
		else:
			var_child2 = -1
		while var_child2 > var_el or var_child1 > var_el:
			if var_child1 > var_child2:
				(self.heap[el], self.heap[child1]) = (self.heap[child1], self.heap[el])
				el = child1
			else:
				(self.heap[el], self.heap[child2]) = (self.heap[child2], self.heap[el])
				el = child2
			var_el = self.mult(el)
			child1 = el * 2
			if child1 < len(self.heap):
				var_child1 = self.mult(child1)
			else:
				var_child1 = -1
			child2 = el * 2 + 1
			if child2 < len(self.heap):
				var_child2 = self.mult(child2)
			else:
				var_child2 = -1
		return ans
(n, m, r, k) = map(int, input().split())
max_x = min(r, n - r + 1)
max_y = min(r, m - r + 1)
col_max_x = n - 2 * (max_x - 1)
col_max_y = m - 2 * (max_y - 1)
Nr = (n - r + 1) * (m - r + 1)
tuple_heap = TupleHeap()
tuple_heap.add((max_x, max_y))
used = {(max_x, max_y)}
SUMM = 0
curr = tuple_heap.pop()
while curr is not None and curr[0] > 0 and (curr[1] > 0) and (k > 0):
	new_tuple1 = (curr[0] - 1, curr[1])
	if new_tuple1 not in used:
		used.add(new_tuple1)
		tuple_heap.add(new_tuple1)
	new_tuple2 = (curr[0], curr[1] - 1)
	if new_tuple2 not in used:
		used.add(new_tuple2)
		tuple_heap.add(new_tuple2)
	x = 2 if curr[0] != max_x else col_max_x
	y = 2 if curr[1] != max_y else col_max_y
	col = min(x * y, k)
	k = k - col
	SUMM = SUMM + col * curr[0] * curr[1]
	curr = tuple_heap.pop()
print(SUMM / Nr)


from heapq import *
(n, m, r, k) = map(int, input().split())
(u, v) = (n // 2, m // 2)
h = []
g = lambda z, l: min(z + 1, l - z, l - r + 1, r)

def f(x, y):
	if 0 <= x < n and 0 <= y < m:
		s = g(x, n) * g(y, m)
		heappush(h, (-s, x, y))
f(u, v)
t = 0
for i in range(k):
	(s, x, y) = heappop(h)
	t -= s
	if x <= u:
		f(x - 1, y)
	if x == u and y <= v:
		f(x, y - 1)
	if x >= u:
		f(x + 1, y)
	if x == u and y >= v:
		f(x, y + 1)
print(t / (n - r + 1) / (m - r + 1))


(n, m, r, k) = map(int, input().split())

def count(y, x):
	minx = max(0, x - (r - 1))
	maxx = min(m - 1 - (r - 1), x)
	miny = max(0, y - (r - 1))
	maxy = min(n - 1 - (r - 1), y)
	res = (maxy - miny + 1) * (maxx - minx + 1)
	return res
(sy, sx) = (n // 2, m // 2)
c = count(sy, sx)
cnt = 0
import heapq
q = [(-c, sy, sx)]
heapq.heapify(q)
visit = set()
visit.add((sy, sx))
E = 0
while q:
	(c, y, x) = heapq.heappop(q)
	c = -c
	E += c
	cnt += 1
	if cnt == k:
		break
	for (dy, dx) in ((-1, 0), (1, 0), (0, -1), (0, 1)):
		(ny, nx) = (y + dy, x + dx)
		if 0 <= ny < n and 0 <= nx < m:
			if (ny, nx) in visit:
				continue
			c = count(ny, nx)
			heapq.heappush(q, (-c, ny, nx))
			visit.add((ny, nx))
ans = E / ((n - r + 1) * (m - r + 1))
print(ans)