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
ii = lambda: sys.stdin.readline().strip()
idata = lambda: [int(x) for x in ii().split()]
sdata = lambda: list(ii())

def solve():
    n, x, y, c = idata()
    r = n ** 2
    l = -1
    while l + 1 < r:
        middle = (l + r) // 2
        ans = 1 + 2 * (middle + 1) * middle
        ans -= pow(middle - n + y, 2) if middle - n + y > 0 else 0
        ans -= pow(middle + 1 - y, 2) if middle + 1 - y > 0 else 0
        ans -= pow(middle - n + x, 2) if middle - n + x > 0 else 0
        ans -= pow(middle + 1 - x, 2) if middle + 1 - x > 0 else 0
        d = 1 + n - x + y
        if middle >= d:
            ans += (middle - d + 1) * (middle - d + 2) // 2
        d = 2 + 2 * n - y - x
        if middle >= d:
            ans += (middle - d + 1) * (middle - d + 2) // 2
        d = 1 + n - y + x
        if middle >= d:
            ans += (middle - d + 1) * (middle - d + 2) // 2
        d = x + y
        if middle >= d:
            ans += (middle - d + 1) * (middle - d + 2) // 2
        if ans >= c:
            r = middle
        else:
            l = middle
    print(r)
    return

for t in range(1):
    solve()