#include <bits/stdc++.h>
using namespace std;
long long side(long long d) {
  if (d <= 0) {
    return 0;
  }
  return d * d;
}
long long corn(long long d) {
  if (d <= 0) {
    return 0;
  }
  return d * (d + 1) / 2;
}
int main() {
  cout.precision(10);
  cout << fixed;
  long long N, X, Y, C;
  cin >> N >> X >> Y >> C;
  --X;
  --Y;
  long long l = -1, r = 2 * N;
  while (r - l > 1) {
    long long m = (l + r) / 2;
    long long q = 2 * m * (m + 1) + 1;
    q -= side(m - X) + side(m - (N - X - 1)) + side(m - Y) +
         side(m - (N - Y - 1));
    q += corn(m - (X + Y + 1)) + corn(m - (X + (N - Y - 1) + 1)) +
         corn(m - ((N - X - 1) + Y + 1)) +
         corn(m - ((N - X - 1) + (N - Y - 1) + 1));
    if (q >= C) {
      r = m;
    } else {
      l = m;
    }
  }
  cout << r << '\n';
  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 side(long long d) {
  if (d <= 0) {
    return 0;
  }
  return d * d;
}
long long corn(long long d) {
  if (d <= 0) {
    return 0;
  }
  return d * (d + 1) / 2;
}
int main() {
  cout.precision(10);
  cout << fixed;
  long long N, X, Y, C;
  cin >> N >> X >> Y >> C;
  --X;
  --Y;
  long long l = -1, r = 2 * N;
  while (r - l > 1) {
    long long m = (l + r) / 2;
    long long q = 2 * m * (m + 1) + 1;
    q -= side(m - X) + side(m - (N - X - 1)) + side(m - Y) +
         side(m - (N - Y - 1));
    q += corn(m - (X + Y + 1)) + corn(m - (X + (N - Y - 1) + 1)) +
         corn(m - ((N - X - 1) + Y + 1)) +
         corn(m - ((N - X - 1) + (N - Y - 1) + 1));
    if (q >= C) {
      r = m;
    } else {
      l = m;
    }
  }
  cout << r << '\n';
  return 0;
}

```