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;
import java.util.StringTokenizer;

public class D {

	static long x,y,n;
	public static void main(String[] args) throws IOException {
		InputReader in = new InputReader();
		n = in.nextLong();
		x = in.nextLong();
		y = in.nextLong();
		long c = in.nextLong();
		long lo = 0;
		long hi = 1000000001;
		while(lo<hi){
			long mid = (lo+hi)/2;
//			System.out.println(getNum(mid)+" "+mid);
			if(getNum(mid)<c) lo = mid+1;
			else hi = mid;
		}
		System.out.println(lo);
	}
	public static long getNum(long s) {
		long total = 2*s*s+2*s+1;
		long r = (s-(n-x))*(s-(n-x));
		if(s<=n-x)r=0;
		long l = (s-(x-1))*(s-(x-1));
		if(s<=x-1)l=0;
		long d = (s-(n-y))*(s-(n-y));
		if(s<=n-y)
			d=0;
		long u = (s-(y-1))*(s-(y-1));
		if(s<=y-1)
			u = 0;
		long rd = (((s-(n-x)-1)-(n-y))*(((s-(n-x)-1)-(n-y))+1))/2;
		if(((s-(n-x)-1)-(n-y))<0)rd=0;
		long ru = (((s-(n-x)-1)-(y-1))*((s-(n-x)-1)-(y-1)+1))/2;
		if((s-(n-x)-1)-(y-1)<0)ru=0;
		long ld = (((s-(x-1)-1)-(n-y))*(((s-(x-1)-1)-(n-y))+1))/2;
		if((s-(x-1)-1)-(n-y)<0)ld=0;
		long lu = (((s-(x-1)-1)-(y-1))*((s-(x-1)-1)-(y-1)+1))/2;
		if((s-(x-1)-1)-(y-1)<0)lu=0;
		
		return total-r-l-d-u+rd+ru+ld+lu;
	}

	static class InputReader {
		BufferedReader in;
		StringTokenizer st;

		
		
		
		public InputReader() throws IOException {
			in = new BufferedReader(new InputStreamReader(System.in));
			st = new StringTokenizer(in.readLine());
		}

		public double nextDouble() throws IOException {
			return Double.parseDouble(next());
		}

		public String next() throws IOException {
			while (!st.hasMoreElements())
				st = new StringTokenizer(in.readLine());
			return st.nextToken();
		}

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

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