#include <bits/stdc++.h>
using namespace std;
int main() {
  long long n, x, y, c, l = 0, r = 2000000000, m, s;
  cin >> n >> x >> y >> c;
  while (l < r) {
    m = l + (r - l) / 2;
    s = 2 * m * (m + 1) + 1;
    s -= max(0ll, x + m - n) * max(0ll, x + m - n);
    s -= max(0ll, y + m - n) * max(0ll, y + m - n);
    s -= max(0ll, m - x + 1) * max(0ll, m - x + 1);
    s -= max(0ll, m - y + 1) * max(0ll, m - y + 1);
    s +=
        max(0ll, m + (x - n + y - n)) * (max(0ll, m + (x - n + y - n)) - 1) / 2;
    s +=
        max(0ll, m + (x - n - y + 1)) * (max(0ll, m + (x - n - y + 1)) - 1) / 2;
    s += max(0ll, m + (-x + 1 + y - n)) * (max(0ll, m + (-x + 1 + y - n)) - 1) /
         2;
    s += max(0ll, m + (-x + 1 - y + 1)) * (max(0ll, m + (-x + 1 - y + 1)) - 1) /
         2;
    if (s < c)
      l = m + 1;
    else
      r = m;
  }
  cout << r;
  return 0;
}

### Prompt

In cpp, your task is to solve 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;
int main() {
  long long n, x, y, c, l = 0, r = 2000000000, m, s;
  cin >> n >> x >> y >> c;
  while (l < r) {
    m = l + (r - l) / 2;
    s = 2 * m * (m + 1) + 1;
    s -= max(0ll, x + m - n) * max(0ll, x + m - n);
    s -= max(0ll, y + m - n) * max(0ll, y + m - n);
    s -= max(0ll, m - x + 1) * max(0ll, m - x + 1);
    s -= max(0ll, m - y + 1) * max(0ll, m - y + 1);
    s +=
        max(0ll, m + (x - n + y - n)) * (max(0ll, m + (x - n + y - n)) - 1) / 2;
    s +=
        max(0ll, m + (x - n - y + 1)) * (max(0ll, m + (x - n - y + 1)) - 1) / 2;
    s += max(0ll, m + (-x + 1 + y - n)) * (max(0ll, m + (-x + 1 + y - n)) - 1) /
         2;
    s += max(0ll, m + (-x + 1 - y + 1)) * (max(0ll, m + (-x + 1 - y + 1)) - 1) /
         2;
    if (s < c)
      l = m + 1;
    else
      r = m;
  }
  cout << r;
  return 0;
}

```