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>.
from sys import stdin
n,x,y,c=map(int,stdin.readline().split())
x-=1;y-=1;n-=1

def square(x):
    return 0 if x<0 else x**2
def triangle(x):
    return 0 if x<0 else x*(x+1)/2
def f(s):
    return (1+2*s*(s+1)-square(x+s-n)
                       -square(s-x)
                       -square(y+s-n)
                       -square(s-y)
                       +triangle(s+x+y-2*n-1)
                       +triangle(s-x-y-1)
                       +triangle(s+x-y-n-1)
                       +triangle(s+y-x-n-1))
                      
lo=0
hi=2*n
while lo!=hi:
    mid=(lo+hi)/2
    if f(mid)<c:
        lo=mid+1
    else:
        hi=mid
print lo
