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.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;

public class Main {
	static int n,x,y;
	static long c;
public static void main(String[] args) throws IOException {
		
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		String[] split = br.readLine().split(" ");
		n = new Integer(split[0]);
		 x = new Integer(split[1]);
		 y = new Integer(split[2]);
		 c = new Long(split[3]);

		 if(c==1){
			 System.out.println(0);
		 } else {
			 System.out.println(find(0,1000000000));
		 }
		
		
	}

	public static long sumOdd(int i) {
		if(i<=0)
			return 0;
		int n= (i-1)*2+1;		
		long h = n/2;
		long r = 1L*n*(n+1)/2 -h*(h+1);
		return r;
		
	}
	
	public static int find( int min, int max){
		if(min == max)
			return min;
		
		int mid = (min+max)/2;
		long currentC = compute(mid);
		if( currentC>=c) {
			max = mid;	
		} else{
			min = mid+1;
		} 
		return find(min, max);
		
	}
	public static long sumSeries(int n){
		if(n<0)
			return 0;
		return 1L*n*(n+1)/2;
	}
	public static long  compute(int t){
		long r = 1L * t * (t+1) * 2 +1;
		if(x+t>n)
			r-= sumOdd(x+t-n);
		if(x<= t)
			r-=sumOdd(t+1-x);
		if(y+t>n)
			r-=sumOdd(y+t-n);
		if(y<=t)
			r-=sumOdd(t+1-y);
		
		if(x<=t && y<=t ){
	 		r+= sumSeries((t-y+1)-x);
		}
		
		if(x<=t && y+t>n){
			r+= sumSeries((t+y-n)-x);
		}
		if(x+t>n && y<=t) {
			r+= sumSeries((t+x-n)-y);
		}
		if(x+t>n && y+t>n) {
			r+= sumSeries((t+y-n)-(n+1-x));
		}
		
		
		return r;
		
	}
}
