#include <bits/stdc++.h>
using namespace std;
long long n, x, y, c;
long long cal1(long long, long long);
long long cal2(long long, long long);
int main() {
  cin >> n >> x >> y >> c;
  long long lo, hi;
  long long res;
  lo = 0;
  hi = 2 * n - 2;
  while (lo <= hi) {
    long long t = (lo + hi) / 2;
    long long s = t * t + (t + 1) * (t + 1);
    s = s - cal1(n + 1, x + t) - cal1(x - t, 0) - cal1(n + 1, y + t) -
        cal1(y - t, 0);
    s = s + cal2(x - t - y, -n - 1);
    s = s + cal2(x - t + y, 0);
    s = s + cal2(n + 1, x + t - y);
    s = s + cal2(2 * n + 2, x + t + y);
    if (s >= c) {
      res = t;
      hi = t - 1;
    } else {
      lo = t + 1;
    }
  }
  cout << res << endl;
  return 0;
}
long long cal1(long long lo, long long hi) {
  if (lo > hi) return 0;
  long long m = hi - lo + 1;
  return m * m;
}
long long cal2(long long lo, long long hi) {
  if (lo > hi) return 0;
  long long m = hi - lo + 1;
  return (1 + m) * m / 2;
}

### 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 n, x, y, c;
long long cal1(long long, long long);
long long cal2(long long, long long);
int main() {
  cin >> n >> x >> y >> c;
  long long lo, hi;
  long long res;
  lo = 0;
  hi = 2 * n - 2;
  while (lo <= hi) {
    long long t = (lo + hi) / 2;
    long long s = t * t + (t + 1) * (t + 1);
    s = s - cal1(n + 1, x + t) - cal1(x - t, 0) - cal1(n + 1, y + t) -
        cal1(y - t, 0);
    s = s + cal2(x - t - y, -n - 1);
    s = s + cal2(x - t + y, 0);
    s = s + cal2(n + 1, x + t - y);
    s = s + cal2(2 * n + 2, x + t + y);
    if (s >= c) {
      res = t;
      hi = t - 1;
    } else {
      lo = t + 1;
    }
  }
  cout << res << endl;
  return 0;
}
long long cal1(long long lo, long long hi) {
  if (lo > hi) return 0;
  long long m = hi - lo + 1;
  return m * m;
}
long long cal2(long long lo, long long hi) {
  if (lo > hi) return 0;
  long long m = hi - lo + 1;
  return (1 + m) * m / 2;
}

```