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 D {
    private static int n,x,y,c;
    
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        n = sc.nextInt(); x = sc.nextInt(); y = sc.nextInt(); c = sc.nextInt();
        
        long res = 2*n, min=0, max=2*n;
        while(min <= max){
            long mid = (min+max) / 2;
            if (count(mid) >= c) {
                res = mid;
                max = mid-1;
            }
            else min = mid + 1;
        }
        System.out.println(res);
    }
    
    private static long sizeTriangle(long n){
        if(n < 1){
            return 0;
        }
        return n*(n+1)/2;
    }
    
    private static long count(long t){
        long
            left = x - t,
            right = x+t,
            top = y - t,
            bottom = y + t;
        if(left < 1) left = 1;
        if(right > n) right = n;
        if(top < 1) top = 1;
        if(bottom > n) bottom = n;
        
        long sqSize = (bottom-top+1)*(right-left+1);
        long topLeft = sizeTriangle(Math.max(-(t - (y-top) - (x-left)), 0));
        long bottomLeft = sizeTriangle(Math.max(-(t-(bottom-y)-(x-left)), 0));
        long topRight = sizeTriangle(Math.max(-(t-(y-top)-(right-x)), 0));
        long bottomRight = sizeTriangle(Math.max(-(t-(bottom-y)-(right-x)), 0));
        
        return sqSize - topLeft - topRight - bottomLeft - bottomRight;
    }
}
