#include <bits/stdc++.h>
using namespace std;
long long l = 0, r = 500007ll;
long long n, x, y, q, Q;
long long k;
long long f(long long t) {
  long long i, j, J;
  long long s = 0ll;
  long long _min = max(1ll, x - 500007ll);
  long long _max = min(n, x + 500007ll);
  for (i = _min; i <= _max; i++) {
    if (abs(i - x) <= t) {
      j = max(1ll, y - abs(t - abs(i - x)));
      J = min(n, y + abs(t - abs(i - x)));
      s += J - j + 1ll;
    }
  }
  return s;
}
int main() {
  int m;
  cin >> n >> x >> y >> k;
  while (l < r) {
    q = (l + r) / 2ll;
    Q = f(q);
    if (Q < k)
      l = q + 1ll;
    else
      r = q;
  }
  cout << l << endl;
}

### 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 l = 0, r = 500007ll;
long long n, x, y, q, Q;
long long k;
long long f(long long t) {
  long long i, j, J;
  long long s = 0ll;
  long long _min = max(1ll, x - 500007ll);
  long long _max = min(n, x + 500007ll);
  for (i = _min; i <= _max; i++) {
    if (abs(i - x) <= t) {
      j = max(1ll, y - abs(t - abs(i - x)));
      J = min(n, y + abs(t - abs(i - x)));
      s += J - j + 1ll;
    }
  }
  return s;
}
int main() {
  int m;
  cin >> n >> x >> y >> k;
  while (l < r) {
    q = (l + r) / 2ll;
    Q = f(q);
    if (Q < k)
      l = q + 1ll;
    else
      r = q;
  }
  cout << l << endl;
}

```