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: interview

Starter code:

Solutions:
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)
#for i in range(10):
#	print(i, cnt(i))
ini, fin = 0, int(1e9)
cont = int(1e9)
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()