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>.
def q(v):
    return 0 if v < 0 else (v * (v + 1)) / 2
def u(v):
    return 0 if v<0 else v

n, x, y, c = map(int, raw_input().split())
l = 0
r = 1000000000000000000
res=1000000000000000000
while l <= r:
    k = (l + r) / 2
    w = -(y - k - 1)
    e = -(x - k - 1)
    t=y+k-n
    z=x+k-n
    a = (k + 1) * (k + 1) + k * k
    a-=2 * q(w)-u(w) + 2 *q(e)-q(e-y)-u(e)
    a-=2*q(t)-q(t-x)-u(t)+2*q(z)-q(z-y)-q(z-(n-y+1))-u(z)
    if a>=c:
        res=min(res,k)
        r=k-1
    else:
        l=k+1
print res