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.util.Scanner;


public class D156div2 {

	static long ge(long p) {
		return p*2 - 1;
	}
	
	static long gv(long p) {
		return p;
	}
	/**
	 * @param args
	 */
	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		long n = sc.nextLong();
		long x = sc.nextLong();
		long y = sc.nextLong();
		long c = sc.nextLong();
		
		long v = 1;
		long t = 0;
		while (v < c) {
			++t;
			v += 4*t;
			if (x <= t) {
				v -= ge(t-x+1);
			}
			if (y <= t) {
				v -= ge(t-y+1);
			}
			if (n-x < t) {
				v -= ge(t-(n-x));
			}
			if (n-y < t) {
				v -= ge(t-(n-y));
			}
			if (x + y <= t) {
				v += gv(t-x-y+1);
			}
			if (x + n-y+1 <= t) {
				v += gv(t-x-(n-y+1)+1);
			}
			if (n-x+1 + y <= t) {
				v += gv(t-(n-x+1)-y+1);
			}
			if (n-x+1 + n-y+1 <= t) {
				v += gv(t-(n-x+1)-(n-y+1)+1);
			}
		}
		System.out.println(t);
	}

}
