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())
ans = add = 1
while ans < c:
    l1 = add + 1
    l2 = l3 = l4 = l1
    sub = 4
    if x - add < 1:
        t = add - x + 1
        l1 -= t
        l2 -= t
        sub -= 1
    if x + add > n:
        t = add + x - n
        l3 -= t
        l4 -= t
        sub -= 1
    if y - add < 1:
        t = add - y + 1
        l1 -= t
        l4 -= t
        sub -= 1
    if y + add > n:
        t = add + y - n
        l2 -= t
        l3 -= t
        sub -= 1
    if l1 > 0:
        ans += l1
    if l2 > 0:
        ans += l2
    if l3 > 0:
        ans += l3
    if l4 > 0:
        ans += l4
    ans -= sub;
    add += 1
print add - 1