#include <bits/stdc++.h>
using namespace std;
const int INF = 20050226;
long long n, c, x, y;
bool check(long long val) {
  long long total = 1 + 2ll * val * (val + 1);
  total -= min(x - 1 - val, 0ll) * min(x - 1 - val, 0ll);
  total -= min(y - 1 - val, 0ll) * min(y - 1 - val, 0ll);
  total -= min(n - (x + val), 0ll) * min(n - (x + val), 0ll);
  total -= min(n - (y + val), 0ll) * min(n - (y + val), 0ll);
  total += min(x - 1 + y - 1 - val + 1, 0ll) *
           (min(x - 1 + y - 1 - val + 1, 0ll) - 1) / 2;
  total += min(x - 1 + n - y + -val + 1, 0ll) *
           (min(x - 1 + n - y - val + 1, 0ll) - 1) / 2;
  total += min(n - x + n - y - val + 1, 0ll) *
           (min(n - x + n - y - val + 1, 0ll) - 1) / 2;
  total += min(n - x + y - 1 - val + 1, 0ll) *
           (min(n - x + y - 1 - val + 1, 0ll) - 1) / 2;
  return total >= c;
}
int main() {
  cin >> n >> x >> y >> c;
  long long l = -1, r = (long long)INF;
  while (l + 1 < r) {
    long long mid = (l + r) / 2;
    if (check(mid))
      r = mid;
    else
      l = mid;
  }
  cout << r;
}

### 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;
const int INF = 20050226;
long long n, c, x, y;
bool check(long long val) {
  long long total = 1 + 2ll * val * (val + 1);
  total -= min(x - 1 - val, 0ll) * min(x - 1 - val, 0ll);
  total -= min(y - 1 - val, 0ll) * min(y - 1 - val, 0ll);
  total -= min(n - (x + val), 0ll) * min(n - (x + val), 0ll);
  total -= min(n - (y + val), 0ll) * min(n - (y + val), 0ll);
  total += min(x - 1 + y - 1 - val + 1, 0ll) *
           (min(x - 1 + y - 1 - val + 1, 0ll) - 1) / 2;
  total += min(x - 1 + n - y + -val + 1, 0ll) *
           (min(x - 1 + n - y - val + 1, 0ll) - 1) / 2;
  total += min(n - x + n - y - val + 1, 0ll) *
           (min(n - x + n - y - val + 1, 0ll) - 1) / 2;
  total += min(n - x + y - 1 - val + 1, 0ll) *
           (min(n - x + y - 1 - val + 1, 0ll) - 1) / 2;
  return total >= c;
}
int main() {
  cin >> n >> x >> y >> c;
  long long l = -1, r = (long long)INF;
  while (l + 1 < r) {
    long long mid = (l + r) / 2;
    if (check(mid))
      r = mid;
    else
      l = mid;
  }
  cout << r;
}

```