#include <bits/stdc++.h>
using namespace std;
long long Calc(long long X, long long Y, long long N, long long T) {
  long long Ans = 0;
  if (X - T < 1) Ans -= (X - T - 1) * (X - T - 1);
  if (X + T > N) Ans -= (X + T - N) * (X + T - N);
  if (Y - T < 1) Ans -= (Y - T - 1) * (Y - T - 1);
  if (Y + T > N) Ans -= (Y + T - N) * (Y + T - N);
  Ans += (T + 1) * (T + 1) + T * T;
  if ((X - 1) + (Y - 1) < T) {
    long long L = T - (X - 1) - (Y - 1);
    Ans += L * (L - 1) / 2;
  }
  if ((X - 1) + (N - Y) < T) {
    long long L = T - (X - 1) - (N - Y);
    Ans += L * (L - 1) / 2;
  }
  if ((N - X) + (Y - 1) < T) {
    long long L = T - (N - X) - (Y - 1);
    Ans += L * (L - 1) / 2;
  }
  if ((N - X) + (N - Y) < T) {
    long long L = T - (N - X) - (N - Y);
    Ans += L * (L - 1) / 2;
  }
  return Ans;
}
int main() {
  long long N, X, Y, Size;
  cin >> N >> X >> Y >> Size;
  long long Left = 0, Right = N * 2;
  while (Left < Right) {
    long long Mid = (Left + Right) / 2;
    if (Calc(X, Y, N, Mid) < Size)
      Left = Mid + 1;
    else
      Right = Mid;
  }
  cout << Left << endl;
  return 0;
}

### Prompt

Please provide a cpp coded solution to the problem described below:

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 Calc(long long X, long long Y, long long N, long long T) {
  long long Ans = 0;
  if (X - T < 1) Ans -= (X - T - 1) * (X - T - 1);
  if (X + T > N) Ans -= (X + T - N) * (X + T - N);
  if (Y - T < 1) Ans -= (Y - T - 1) * (Y - T - 1);
  if (Y + T > N) Ans -= (Y + T - N) * (Y + T - N);
  Ans += (T + 1) * (T + 1) + T * T;
  if ((X - 1) + (Y - 1) < T) {
    long long L = T - (X - 1) - (Y - 1);
    Ans += L * (L - 1) / 2;
  }
  if ((X - 1) + (N - Y) < T) {
    long long L = T - (X - 1) - (N - Y);
    Ans += L * (L - 1) / 2;
  }
  if ((N - X) + (Y - 1) < T) {
    long long L = T - (N - X) - (Y - 1);
    Ans += L * (L - 1) / 2;
  }
  if ((N - X) + (N - Y) < T) {
    long long L = T - (N - X) - (N - Y);
    Ans += L * (L - 1) / 2;
  }
  return Ans;
}
int main() {
  long long N, X, Y, Size;
  cin >> N >> X >> Y >> Size;
  long long Left = 0, Right = N * 2;
  while (Left < Right) {
    long long Mid = (Left + Right) / 2;
    if (Calc(X, Y, N, Mid) < Size)
      Left = Mid + 1;
    else
      Right = Mid;
  }
  cout << Left << endl;
  return 0;
}

```