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

### Prompt

Your task is to create 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;
long long area(long long x, long long y, long long t) {
  long long ans = 1 + 2 * t * (t + 1);
  if (t > x - 1) ans -= (t - x + 1) * (t - x + 1);
  if (t > y - 1) ans -= (t - y + 1) * (t - y + 1);
  if (t > n - y) ans -= (t - n + y) * (t - n + y);
  if (t > n - x) ans -= (t - n + x) * (t - n + x);
  if (t > x + y - 1) ans += (t - x - y + 1) * (t - x - y + 2) / 2;
  if (t > x + n - y) ans += (t - x - n + y) * (t - x + 1 - n + y) / 2;
  if (t > n - x + y) ans += (t - n + x - y) * (t - n + x - y + 1) / 2;
  if (t > n - x + n - y + 1)
    ans += (t + x + y - 2 * n - 1) * (t + x + y - 2 * n) / 2;
  return ans;
}
int main() {
  cin >> n >> x >> y >> c;
  long long l = 0, r = 2 * n + 1, ans = -1;
  while (l <= r) {
    long long m = l + (r - l) / 2;
    if (area(x, y, m) >= c) {
      ans = m;
      r = m - 1;
    } else {
      l = m + 1;
    }
  }
  cout << ans;
}

```