#include <bits/stdc++.h>
using namespace std;
long long n, x, y, c;
bool ch(long long m) {
  if (m == 0) return c == 1;
  double all = 1 + (m * (m + 1) / 2) * 4;
  double add = m - x + 1;
  if (add > 0) {
    add--;
    all -= (add * (add + 1) / 2) * 2 + add + 1;
  }
  add = m - y + 1;
  if (add > 0) {
    add--;
    all -= (add * (add + 1) / 2) * 2 + add + 1;
  }
  add = m - (n - y + 1) + 1;
  if (add > 0) {
    add--;
    all -= (add * (add + 1) / 2) * 2 + add + 1;
  }
  add = m - (n - x + 1) + 1;
  if (add > 0) {
    add--;
    all -= (add * (add + 1) / 2) * 2 + add + 1;
  }
  long long dx, dy;
  dx = abs(x - 1);
  dy = abs(y - 1);
  add = m - (dx + dy) - 1;
  if (add > 0) {
    all += add * (add + 1) / 2;
  }
  dx = abs(x - 1);
  dy = abs(n - y);
  add = m - (dx + dy) - 1;
  if (add > 0) {
    all += add * (add + 1) / 2;
  }
  dx = abs(n - x);
  dy = abs(y - 1);
  add = m - (dx + dy) - 1;
  if (add > 0) {
    all += add * (add + 1) / 2;
  }
  dx = abs(n - x);
  dy = abs(n - y);
  add = m - (dx + dy) - 1;
  if (add > 0) {
    all += add * (add + 1) / 2;
  }
  return all >= c;
}
int main() {
  ios::sync_with_stdio(0);
  cin >> n >> x >> y >> c;
  long long l = 0, h = 2e9, res = 0;
  while (l <= h) {
    long long m = (l + h) / 2;
    if (ch(m)) {
      res = m;
      h = m - 1;
    } else
      l = m + 1;
  }
  cout << res << endl;
  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, c;
bool ch(long long m) {
  if (m == 0) return c == 1;
  double all = 1 + (m * (m + 1) / 2) * 4;
  double add = m - x + 1;
  if (add > 0) {
    add--;
    all -= (add * (add + 1) / 2) * 2 + add + 1;
  }
  add = m - y + 1;
  if (add > 0) {
    add--;
    all -= (add * (add + 1) / 2) * 2 + add + 1;
  }
  add = m - (n - y + 1) + 1;
  if (add > 0) {
    add--;
    all -= (add * (add + 1) / 2) * 2 + add + 1;
  }
  add = m - (n - x + 1) + 1;
  if (add > 0) {
    add--;
    all -= (add * (add + 1) / 2) * 2 + add + 1;
  }
  long long dx, dy;
  dx = abs(x - 1);
  dy = abs(y - 1);
  add = m - (dx + dy) - 1;
  if (add > 0) {
    all += add * (add + 1) / 2;
  }
  dx = abs(x - 1);
  dy = abs(n - y);
  add = m - (dx + dy) - 1;
  if (add > 0) {
    all += add * (add + 1) / 2;
  }
  dx = abs(n - x);
  dy = abs(y - 1);
  add = m - (dx + dy) - 1;
  if (add > 0) {
    all += add * (add + 1) / 2;
  }
  dx = abs(n - x);
  dy = abs(n - y);
  add = m - (dx + dy) - 1;
  if (add > 0) {
    all += add * (add + 1) / 2;
  }
  return all >= c;
}
int main() {
  ios::sync_with_stdio(0);
  cin >> n >> x >> y >> c;
  long long l = 0, h = 2e9, res = 0;
  while (l <= h) {
    long long m = (l + h) / 2;
    if (ch(m)) {
      res = m;
      h = m - 1;
    } else
      l = m + 1;
  }
  cout << res << endl;
  return 0;
}

```