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 java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;

public class Square {
	public static void main(String[] args) throws IOException {
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));

		String[] line = br.readLine().split(" ");
		int n = Integer.parseInt(line[0]);
		int x = Integer.parseInt(line[1]);
		int y = Integer.parseInt(line[2]);
		int c = Integer.parseInt(line[3]);

		int[] dist = new int[4];
		dist[0] = x - 1;
		dist[1] = y - 1;
		dist[2] = n - x;
		dist[3] = n - y;

		int total = 1;
		int pasos = 0;
		while (total < c) {

			pasos++;
			total += 4 * pasos;

			for (int i = 0; i < dist.length; i++) {
				if (pasos - dist[i] > 0) {
					total -= (pasos - dist[i]) * 2 - 1;
				}
				if (pasos - dist[i] - dist[(i+1)%4] - 1 > 0){
					total += pasos - dist[i] - dist[(i+1)%4] - 1;
				}
			}
		}

		System.out.println(pasos);
	}
}
