Mr. Bender has a digital table of size n × n, each cell can be switched on or off. He wants the field to have at least c switched on squares. When this condition is fulfilled, Mr Bender will be happy.

We'll consider the table rows numbered from top to bottom from 1 to n, and the columns — numbered from left to right from 1 to n. Initially there is exactly one switched on cell with coordinates (x, y) (x is the row number, y is the column number), and all other cells are switched off. Then each second we switch on the cells that are off but have the side-adjacent cells that are on.

For a cell with coordinates (x, y) the side-adjacent cells are cells with coordinates (x - 1, y), (x + 1, y), (x, y - 1), (x, y + 1).

In how many seconds will Mr. Bender get happy?


-----Input-----

The first line contains four space-separated integers n, x, y, c (1 ≤ n, c ≤ 10^9; 1 ≤ x, y ≤ n; c ≤ n^2).


-----Output-----

In a single line print a single integer — the answer to the problem.


-----Examples-----
Input
6 4 3 1

Output
0

Input
9 3 8 10

Output
2



-----Note-----

Initially the first test has one painted cell, so the answer is 0. In the second test all events will go as is shown on the figure. [Image].

Difficulty: MEDIUM_HARD

Solutions:
def compute(n, x, y, c):
	x1 = min(x - 1, n - x)
	x2 = n - x1 - 1
	y1 = min(y - 1, n - y)
	y2 = n - y1 - 1
	i1 = x1 + y1 + 1
	i2 = min(y1 + x2, y2 + x1) + 1
	i3 = max(y1 + x2, y2 + x1) + 1
	o1 = min(x1, y1)
	o2 = min(max(x1, y1), min(y2, x2))
	o3 = max(max(x1, y1), min(y2, x2))
	o4 = max(y2, x2)
	inside = 0
	out = 0
	s = 0
	current = 1
	Happy = False
	if c == 1:
		print(0)
		return 0
	while Happy == False:
		s += 1
		new = (s + 1) * 4 - 4
		out = out + out_new(s, o1) + out_new(s, o2) + out_new(s, o3) + out_new(s, o4)
		inside = inside + in_new(s, i1) + in_new(s, i2) + in_new(s, i3)
		current = current + new
		test = current - out + inside
		if test >= c:
			print(s)
			break
	return s

def out_new(s, sx):
	if sx < s:
		return (s - sx) * 2 - 1
	else:
		return 0

def in_new(s, sx):
	if sx < s:
		return s - sx
	else:
		return 0
x = input()
y = x.split()
a = int(y[0])
b = int(y[1])
c = int(y[2])
d = int(y[3])
compute(a, b, c, d)


(x, y, n, c) = (0, 0, 0, 0)

def suma_impares(m):
	return m * m

def suma_n(m):
	return m * (m - 1) // 2

def cnt(t):
	(u, d, l, r) = (x + t, x - t, y - t, y + t)
	suma = t ** 2 + (t + 1) ** 2
	if u > n:
		suma -= suma_impares(u - n)
	if d < 1:
		suma -= suma_impares(1 - d)
	if l < 1:
		suma -= suma_impares(1 - l)
	if r > n:
		suma -= suma_impares(r - n)
	if 1 - l > x - 1 and 1 - d > y - 1:
		suma += suma_n(2 - l - x)
	if r - n > x - 1 and 1 - d > n - y:
		suma += suma_n(r - n - x + 1)
	if 1 - l > n - x and u - n > y - 1:
		suma += suma_n(1 - l - n + x)
	if u - n > n - y and r - n > n - x:
		suma += suma_n(u - n - n + y)
	return suma
(n, x, y, c) = input().split()
(n, x, y, c) = (int(n), int(x), int(y), int(c))
(ini, fin) = (0, int(1000000000.0))
cont = int(1000000000.0)
while cont > 0:
	m = ini
	paso = cont // 2
	m += paso
	if cnt(m) < c:
		ini = m + 1
		cont -= paso + 1
	else:
		cont = paso
print(ini)


import sys
ii = lambda : sys.stdin.readline().strip()
idata = lambda : [int(x) for x in ii().split()]
sdata = lambda : list(ii())

def solve():
	(n, x, y, c) = idata()
	r = n ** 2
	l = -1
	while l + 1 < r:
		middle = (l + r) // 2
		ans = 1 + 2 * (middle + 1) * middle
		ans -= pow(middle - n + y, 2) if middle - n + y > 0 else 0
		ans -= pow(middle + 1 - y, 2) if middle + 1 - y > 0 else 0
		ans -= pow(middle - n + x, 2) if middle - n + x > 0 else 0
		ans -= pow(middle + 1 - x, 2) if middle + 1 - x > 0 else 0
		d = 1 + n - x + y
		if middle >= d:
			ans += (middle - d + 1) * (middle - d + 2) // 2
		d = 2 + 2 * n - y - x
		if middle >= d:
			ans += (middle - d + 1) * (middle - d + 2) // 2
		d = 1 + n - y + x
		if middle >= d:
			ans += (middle - d + 1) * (middle - d + 2) // 2
		d = x + y
		if middle >= d:
			ans += (middle - d + 1) * (middle - d + 2) // 2
		if ans >= c:
			r = middle
		else:
			l = middle
	print(r)
	return
for t in range(1):
	solve()