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())
l,h=0,n*3
V=lambda z:z>0 and z*(z+1)/2
W=lambda z:z>0 and z*z
while l<h:
  m=(l+h)/2
  s=V(m)*4+1-W(1-x+m)-W(x+m-n)-W(1-y+m)-W(y+m-n)+V(x+y+m-n-n-1)+V(2-x-y+m-1)+V(x+m-y-n+1-1)+V(1-n-x+m+y-1)
  if s<c:
    l=m+1
  else:
    h=m
print l
