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.*;
import java.util.*;

public class D {
    public static void main(String[] args) throws IOException
    {
        new D().run();
    }

    InputReader in;
    PrintWriter out;

    public void run() throws IOException
    {
        in = new InputReader( System.in );
        out = new PrintWriter( System.out );
        solution();
        out.close();
    }

    /* Start the solution here */
    long n, x, y, c;

    long func(long t)
    {
        long left = Math.min( t, y - 1 );
        long right = Math.min( t, n - y );
        long up = Math.min( t, x - 1 );
        long down = Math.min( t, n - x );

        long ret = 1 + left + right + up + down;
        ret += square(t, left, up);
        ret += square(t, up, right);
        ret += square(t, right, down);
        ret += square(t, down, left);
        return ret;
    }

    long triangular(long n)
    {
        return n * (n+1) / 2;
    }

    private long square(long t, long A, long B)
    {
        if ( A > B )
        {
            long temp = A;
            A = B;
            B = temp;
        }
        if ( t >= A+B ) return A*B;
        if ( t < A ) return triangular( t-1 );
        else return A*B - triangular( (A+B)-t );
    }

    private void solution() throws IOException
    {
        n = in.nextLong();
        x = in.nextLong();
        y = in.nextLong();
        c = in.nextLong();

        int low = 0, high=(int)1e9;
        while ( low < high )
        {
            int mid = (low + high) / 2;
            if ( func( mid ) < c ) low = mid + 1;
            else high = mid;
        }

        out.println( low );
    }
}

class InputReader
{
    BufferedReader br;
    StringTokenizer st;

    public InputReader(InputStream in)
    {
        br = new BufferedReader(new InputStreamReader(in));
        st = null;
    }

    public String next() throws IOException
    {
        while (st==null || !st.hasMoreTokens())
        {
            st = new StringTokenizer(br.readLine());
        }
        return st.nextToken();
    }

    public int nextInt() throws IOException
    {
        return Integer.parseInt(next());
    }

    public long nextLong() throws IOException
    {
        return Long.parseLong( next() );
    }
}