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;
long long n, x, y, c, res, S;
int main() {
  cin >> n >> x >> y >> c;
  long long dau = 0, cuoi = 707106781, giua;
  while (dau + 1 < cuoi) {
    giua = (dau + cuoi) / 2;
    if (2 * (giua - 1) * (giua - 1) + 2 * giua - 1 < c)
      dau = giua;
    else
      cuoi = giua;
  }
  res = cuoi;
  if (res == 1)
    cout << 0 << endl;
  else {
    S = 2 * (res - 1) * (res - 1) + 2 * res - 1;
    long long x1, x2, y1, y2;
    x1 = x;
    y1 = y;
    x2 = n - x + 1;
    y2 = n - y + 1;
    if (res >= x1) S = S - (res - x1) * (res - x1);
    if (res >= x2) S = S - (res - x2) * (res - x2);
    if (res >= y1) S = S - (res - y1) * (res - y1);
    if (res >= y2) S = S - (res - y2) * (res - y2);
    if (res > x1 + y2) S = S + (res - x1 - y2) * (res - x1 - y2 + 1) / 2;
    if (res > x1 + y1) S = S + (res - x1 - y1) * (res - x1 - y1 + 1) / 2;
    if (res > x2 + y1) S = S + (res - x2 - y1) * (res - x2 - y1 + 1) / 2;
    if (res > x2 + y2) S = S + (res - x2 - y2) * (res - x2 - y2 + 1) / 2;
    while (S < c) {
      res++;
      S = 2 * (res - 1) * (res - 1) + 2 * res - 1;
      x1 = x;
      y1 = y;
      x2 = n - x + 1;
      y2 = n - y + 1;
      if (res >= x1) S = S - (res - x1) * (res - x1);
      if (res >= x2) S = S - (res - x2) * (res - x2);
      if (res >= y1) S = S - (res - y1) * (res - y1);
      if (res >= y2) S = S - (res - y2) * (res - y2);
      if (res > x1 + y2) S = S + (res - x1 - y2) * (res - x1 - y2 + 1) / 2;
      if (res > x1 + y1) S = S + (res - x1 - y1) * (res - x1 - y1 + 1) / 2;
      if (res > x2 + y1) S = S + (res - x2 - y1) * (res - x2 - y1 + 1) / 2;
      if (res > x2 + y2) S = S + (res - x2 - y2) * (res - x2 - y2 + 1) / 2;
    }
    cout << res - 1 << endl;
  }
}
