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 ≤ 109; 1 ≤ x, y ≤ n; c ≤ n2).

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>.
def g(v):
	return 0 if v < 0 else v * v

def h(v):
	return 0 if v < 0 else v * (v + 1) / 2

def f(v):
	return g(v + 1) + g(v) - g(v + 1 - x) - g(v + 1 - n - 1 + x) - g(v + 1 - y) - g(v + 1 - n - 1 + y) + h(v + 1 - x - y) + h(v + 1 - n - 1 + x - y) + h(v + 1 - x - n - 1 + y) + h(v + 1 - n - 1 + x - n - 1 + y)

n, x, y, c = map(int, raw_input().split())
l = 0
r = c
while l <= r:
	v = (l + r) / 2
	if f(v) < c:
		l = v + 1
	else:
		r = v - 1
		p = v
print p