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>.
#include <bits/stdc++.h>
using namespace std;
int main() {
  long long n, x, y, g;
  cin >> n >> x >> y >> g;
  long long a = y - 1;
  long long b = n - x;
  long long c = n - y;
  long long d = x - 1;
  long long on = 1;
  long long s = 0;
  while (on < g) {
    s++;
    on += 4 * s;
    if (s > a) on -= (s - a) * 2 - 1;
    if (s > b) on -= (s - b) * 2 - 1;
    if (s > c) on -= (s - c) * 2 - 1;
    if (s > d) on -= (s - d) * 2 - 1;
    if (s > (a + b) + 1) on += (s - (a + b + 1));
    if (s > (b + c) + 1) on += (s - (b + c + 1));
    if (s > (c + d) + 1) on += (s - (c + d + 1));
    if (s > (d + a) + 1) on += (s - (d + a + 1));
  }
  cout << s << endl;
  return 0;
}
