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

sum_n = lambda n: n > 0 and (n * (n + 1)) // 2
sum_odd = lambda n: n > 0 and n * n

n, x, y, c = [int(x) for x in stdin.readline().split()]
ans, cur = 0, 1

while cur < c:
    ans += 1
    up, down, left, right = 1 - x + ans, x + ans - n, 1 - y + ans, y + ans - n

    cur = sum_n(ans) * 4 + 1 - sum_odd(up) - sum_odd(down) - sum_odd(left) - sum_odd(right)
    cur += sum_n(right - x) + sum_n(right - n + x - 1) + sum_n(left - x) + sum_n(left - n + x - 1)

print(ans)
