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

### Prompt

Your challenge is to write a cpp solution to the following problem:

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>.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
  long long n, x, y, c;
  cin >> n >> x >> y >> c;
  long long cnt = 1;
  int t = 1;
  n++;
  while (c > cnt) {
    t++;
    cnt += 4 * (t - 1);
    if (x < t) cnt -= 2 * (t - x) - 1;
    if (y < t) cnt -= 2 * (t - y) - 1;
    if (n - x < t) cnt -= 2 * (t - (n - x)) - 1;
    if (n - y < t) cnt -= 2 * (t - (n - y)) - 1;
    if (x + y < t) cnt += t - (x + y);
    if (x + (n - y) < t) cnt += t - (x + n - y);
    if (n - x + y < t) cnt += t - (n - x + y);
    if (n - x + (n - y) < t) cnt += t - (n - x + n - y);
  }
  cout << (t - 1) << endl;
  return 0;
}

```