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>.
// practice with rainboy
import java.io.*;
import java.util.*;

public class CF256B extends PrintWriter {
	CF256B() { super(System.out, true); }
	Scanner sc = new Scanner(System.in);
	public static void main(String[] $) {
		CF256B o = new CF256B(); o.main(); o.flush();
	}

	long count(int n, int x, int y, int r) {
		long a = 2L * r * r + 2 * r + 1;
		int w;
		if ((w = 1 - (x - r)) > 0)
			a -= (long) w * w;
		if ((w = 1 - (y - r)) > 0)
			a -= (long) w * w;
		if ((w = (x + r) - n) > 0)
			a -= (long) w * w;
		if ((w = (y + r) - n) > 0)
			a -= (long) w * w;
		if ((w = r - 1 - (x - 1) - (y - 1)) > 0)
			a += (long) w * (w + 1) / 2;
		if ((w = r - 1 - (x - 1) - (n - y)) > 0)
			a += (long) w * (w + 1) / 2;
		if ((w = r - 1 - (n - x) - (y - 1)) > 0)
			a += (long) w * (w + 1) / 2;
		if ((w = r - 1 - (n - x) - (n - y)) > 0)
			a += (long) w * (w + 1) / 2;
		return a;
	}
	void main() {
		int n = sc.nextInt();
		int x = sc.nextInt();
		int y = sc.nextInt();
		int c = sc.nextInt();
		int lower = -1, upper = c;
		while (upper - lower > 1) {
			int r = (lower + upper) / 2;
			if (count(n, x, y, r) >= c)
				upper = r;
			else
				lower = r;
		}
		println(upper);
	}
}
