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

### Prompt

Create a solution in CPP for 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;
long long n, x, y, spd, c;
int main() {
  cin >> n >> x >> y >> c;
  for (long long k = 0; spd < n * n; k++) {
    spd = 2 * (k + 1) * k + 1;
    if (k >= x) spd -= (k - x + 1) * (k - x + 1);
    if (k > n - x) spd -= (k - n + x) * (k - n + x);
    if (k >= y) spd -= (k - y + 1) * (k - y + 1);
    if (k > n - y) spd -= (k - n + y) * (k - n + y);
    if (x + y + k > n * 2 + 1)
      spd += (x + y + k - n * 2 - 1) * (x + y + k - n * 2) / 2;
    if (x + k - n - y + 1 > 1)
      spd += (x + k - n - y + 1 - 1) * (x + k - n - y + 1) / 2;
    if (y + k - n - x + 1 > 1)
      spd += (y + k - n - x + 1 - 1) * (y + k - n - x + 1) / 2;
    if (k - x + 1 - y + 1 > 1)
      spd += (k - x + 1 - y + 1 - 1) * (k - x + 1 - y + 1) / 2;
    if (spd >= c) {
      cout << k;
      return 0;
    }
  }
  return 0;
}

```