#include <bits/stdc++.h>
using namespace std;
int n, x, y, c;
bool works(int d) {
  long long int _c = c;
  for (int dx = 0; x - dx >= 1 && dx <= d && _c > 0; ++dx) {
    int lim_dy = d - dx;
    _c -= min(n - y, lim_dy) + 1LL;
    _c -= min(y - 1, lim_dy) + 1LL;
    ++_c;
  }
  for (int dx = 1; x + dx <= n && dx <= d && _c > 0; ++dx) {
    int lim_dy = d - dx;
    _c -= min(n - y, lim_dy) + 1LL;
    _c -= min(y - 1, lim_dy) + 1LL;
    ++_c;
  }
  return _c <= 0;
}
int main() {
  cin >> n >> x >> y >> c;
  int st = 0;
  int dr = 200000;
  int mid;
  int ans = dr + 1;
  while (st <= dr) {
    mid = (st + dr) >> 1;
    if (works(mid)) {
      ans = mid;
      dr = mid - 1;
    } else
      st = mid + 1;
  }
  cout << ans << '\n';
  return 0;
}

### Prompt

Your task is to create 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 n, x, y, c;
bool works(int d) {
  long long int _c = c;
  for (int dx = 0; x - dx >= 1 && dx <= d && _c > 0; ++dx) {
    int lim_dy = d - dx;
    _c -= min(n - y, lim_dy) + 1LL;
    _c -= min(y - 1, lim_dy) + 1LL;
    ++_c;
  }
  for (int dx = 1; x + dx <= n && dx <= d && _c > 0; ++dx) {
    int lim_dy = d - dx;
    _c -= min(n - y, lim_dy) + 1LL;
    _c -= min(y - 1, lim_dy) + 1LL;
    ++_c;
  }
  return _c <= 0;
}
int main() {
  cin >> n >> x >> y >> c;
  int st = 0;
  int dr = 200000;
  int mid;
  int ans = dr + 1;
  while (st <= dr) {
    mid = (st + dr) >> 1;
    if (works(mid)) {
      ans = mid;
      dr = mid - 1;
    } else
      st = mid + 1;
  }
  cout << ans << '\n';
  return 0;
}

```