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>.
import sys

def sum(i):
    if i%2 == 0:
        return (i+1)*(i/2)
    else:
        return (i+1)*((i-1)/2)+((i+1)/2)

def switched(i,m,k):
    if i <= min(m,k):
        return sum(i)
    elif i>min(m,k) and i<max(m,k):
        return sum(min(m,k)) + (min(m,k)*(i-min(m,k)))
    elif i>=max(m,k) and i < max(m,k)+min(m,k):
        tmp = i - max(m,k)+1
        return sum(min(m,k)) + (min(m,k)*(max(m,k)-min(m,k)-1)) + sum(min(m,k))-sum(min(m,k)-tmp)
    else:
        return m*k
        
if __name__ == '__main__':
    n,x,y,c = map(int,raw_input().split())
    if c == 1:
        print 0
    else:
        i=1
        num = 1
        while num < c-1:
            num = switched(i,x,y-1) + switched(i,n-x,y) + switched(i,n-x+1,n-y) + switched(i,x-1,n-y+1)
            i = i + 1
        print i - 1