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.*;
import java.io.*;

public class MrBenderAndSquare {
	public static void main(String[] args) throws IOException {
		BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
		PrintWriter out = new PrintWriter(System.out);
		StringTokenizer st = new StringTokenizer(in.readLine());
		long N = Integer.parseInt(st.nextToken());
		long X = Integer.parseInt(st.nextToken());
		long Y = Integer.parseInt(st.nextToken());
		long C = Integer.parseInt(st.nextToken());
		long l = 0;
		long r = 2*N;
		while(l < r) {
			long t = (l + r) / 2;
			long tot = 1 + 2*t + 2*t*t;
			if(t > X - 1) {
				tot -= (t - (X - 1))*(t - (X - 1));
			}
			if(t > Y - 1) {
				tot -= (t - (Y - 1))*(t - (Y - 1));
			}
			if(X + t > N) {
				tot -= (X + t - N)*(X + t - N);
			}
			if(Y + t > N) {
				tot -= (Y + t - N)*(Y + t - N);
			}
			if(t > X + Y - 1) {
				tot += (t - (X + Y - 1))*(t - (X + Y - 1) + 1)/2;
			}
			if(t > (N - X) + Y) {
				tot += (t - ((N - X) + Y))*(t - ((N - X) + Y) + 1)/2;
			}
			if(t > X + (N - Y)) {
				tot += (t - (X + (N - Y)))*(t - (X + (N - Y)) + 1)/2;
			}
			if(t > (N - X) + (N - Y) + 1) {
				tot += (t - ((N - X) + (N - Y) + 1))*(t - ((N - X) + (N - Y) + 1) + 1)/2;
			}
			if(tot < C) {
				l = t + 1;
			} else {
				r = t;
			}
		}
		out.println(l);
		in.close();
		out.close();
	}
}