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>.
n,x,y,c = map(int,raw_input().split())
def count(t):
	t1 = (2*t+1)**2 - 2*t*(t+1)
	if(t>n-x):
		t1-=(t-(n-x))**2
	if(t>n-y):
		t1-=(t-(n-y))**2
	if(t>x-1):
		t1-=(t-(x-1))**2
	if(t>y-1):
		t1-=(t-(y-1))**2

	e1 = (n-x) + (n-y)
	e2 = (n-y) + (x-1)
	e3 = (x-1) + (y-1)
	e4 = (y-1) + (n-x)

	for e in [e1,e2,e3,e4]:
		if(t>e+1):
			p = t-(e+1)
			t1 += (p*(p+1))/2

	# print 't1',t1
	return t1

beg=0
end=n*2
while(beg!=end):
	mid = (beg+end)/2
	if(count(mid)<c):
		beg = mid+1
	else:
		end = mid

print beg	