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.*;

public class MarsLandVaccine{
    public static void main(String[] args){
        Scanner hp = new Scanner(System.in);

        long n = hp.nextLong();
        long x = hp.nextLong();
        long y = hp.nextLong();

        long k = hp.nextLong();

        long low = 0, high = 2 * n;

        while(low < high){
            long mid = (low + high) / 2;
            if(isValidTime(mid, n, k, x, y)){
                high = mid;
            }
            else{
                low = mid + 1;
            }

        }
        System.out.println(low);
    }

    public static boolean isValidTime(long days, long N, long k, long x, long y){

        long top = Math.max(0, (days - (x - 1)));
        long lft = Math.max(0, (days - (y - 1)));
        long rgt = Math.max(0, (y - (N - days)));
        long btm = Math.max(0, (x - (N - days)));

        long topLft = Math.max(0, (1 + days) - (x + y));
        long topRgt = Math.max(0, (1 + days) - (x + N - y + 1));
        long botLft = Math.max(0, (1 + days) - (y + N - x + 1));
        long botRgt = Math.max(0, (1 + days) - (N - x + 1 + N - y + 1));

        long answer = ((days * days) + ((days + 1) * (days + 1)));

        answer -= ((top * top) + (btm * btm) + (lft * lft) + (rgt * rgt));

        answer += (topLft * (topLft + 1)) / 2;
        answer += (topRgt * (topRgt + 1)) / 2;
        answer += (botLft * (botLft + 1)) / 2;
        answer += (botRgt * (botRgt + 1)) / 2;

        return (answer >= k);
    }
}
