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, c;
  cin >> n >> y >> x >> c;
  long long now = 0;
  long long i, xr, xl, yu, yd, d;
  x--;
  y--;
  for (i = 1; now < c; i++) {
    now = i * i + (i - 1) * (i - 1);
    if (now < c) continue;
    i--;
    xr = x + i;
    xl = x - i;
    yu = y - i;
    yd = y + i;
    if (xl < 0) now -= xl * xl;
    if (xr > n - 1) now -= (xr - n + 1) * (xr - n + 1);
    if (yu < 0) {
      now -= yu * yu;
      yu++;
      d = -yu;
      if (x + d > n - 1) now += (x + d - n + 2) * (x + d - n + 1) / 2;
      if (x + yu < 0) now += (x - d) * (x - d - 1) / 2;
    }
    if (yd > n - 1) {
      now -= (yd - n + 1) * (yd - n + 1);
      d = yd - n + 1;
      d--;
      if (x + d > n - 1) now += (x + d - n + 2) * (x + d - n + 1) / 2;
      if (x - d < 0) now += (x - d) * (x - d - 1) / 2;
    }
    i++;
  }
  cout << i - 2 << endl;
}
