#include <bits/stdc++.h>
using namespace std;
long long n, x, y, c, t;
long long two = 2, one = 1;
long long f(long long x, long long y, long long t) {
  long long ans = 0;
  if ((t == 0) || (x * y == 0)) return 0;
  if (x + y <= t) return x * y;
  ans = (one + t) * t / two;
  if (t > x) ans = ans - (one + t - x) * (t - x) / two;
  if (t > y) ans = ans - (one + t - y) * (t - y) / two;
  return ans;
}
int main() {
  cin >> n >> x >> y >> c;
  long long l = 0, r = n + n + n;
  while (l < r) {
    long long mid = (l + r) / two;
    long long tmp = 1;
    tmp += f(x - one, y, mid) + f(n - x + one, y - one, mid) +
           f(n - x, n - y + one, mid) + f(x, n - y, mid);
    if (tmp >= c)
      r = mid;
    else
      l = mid + one;
  }
  cout << l << endl;
  return 0;
}

### Prompt

Generate a Cpp solution to 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, t;
long long two = 2, one = 1;
long long f(long long x, long long y, long long t) {
  long long ans = 0;
  if ((t == 0) || (x * y == 0)) return 0;
  if (x + y <= t) return x * y;
  ans = (one + t) * t / two;
  if (t > x) ans = ans - (one + t - x) * (t - x) / two;
  if (t > y) ans = ans - (one + t - y) * (t - y) / two;
  return ans;
}
int main() {
  cin >> n >> x >> y >> c;
  long long l = 0, r = n + n + n;
  while (l < r) {
    long long mid = (l + r) / two;
    long long tmp = 1;
    tmp += f(x - one, y, mid) + f(n - x + one, y - one, mid) +
           f(n - x, n - y + one, mid) + f(x, n - y, mid);
    if (tmp >= c)
      r = mid;
    else
      l = mid + one;
  }
  cout << l << endl;
  return 0;
}

```