#include <bits/stdc++.h>
using namespace std;
long long int n, x, y;
long long int cnt(long long int d) {
  long long int res = 0;
  long long int xv, yv;
  xv = min(x - 1, d - 1);
  yv = min(y - 1, d - 1);
  res += max(xv + yv - d + 1, 0LL);
  xv = min(n - x, d - 1);
  yv = min(n - y, d - 1);
  res += max(xv + yv - d + 1, 0LL);
  xv = min(n - x, d - 1);
  yv = min(y - 1, d - 1);
  res += max(xv + yv - d + 1, 0LL);
  xv = min(x - 1, d - 1);
  yv = min(n - y, d - 1);
  res += max(xv + yv - d + 1, 0LL);
  res += (x > d);
  res += (y > d);
  res += ((x + d) <= n);
  res += ((y + d) <= n);
  return res;
}
int main() {
  long long int c;
  cin >> n >> x >> y >> c;
  long long int res = 1;
  for (long long int i = 1;; i++) {
    if (res >= c) {
      cout << i - 1 << endl;
      return 0;
    }
    res += cnt(i);
  }
}

### Prompt

Please formulate 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 int n, x, y;
long long int cnt(long long int d) {
  long long int res = 0;
  long long int xv, yv;
  xv = min(x - 1, d - 1);
  yv = min(y - 1, d - 1);
  res += max(xv + yv - d + 1, 0LL);
  xv = min(n - x, d - 1);
  yv = min(n - y, d - 1);
  res += max(xv + yv - d + 1, 0LL);
  xv = min(n - x, d - 1);
  yv = min(y - 1, d - 1);
  res += max(xv + yv - d + 1, 0LL);
  xv = min(x - 1, d - 1);
  yv = min(n - y, d - 1);
  res += max(xv + yv - d + 1, 0LL);
  res += (x > d);
  res += (y > d);
  res += ((x + d) <= n);
  res += ((y + d) <= n);
  return res;
}
int main() {
  long long int c;
  cin >> n >> x >> y >> c;
  long long int res = 1;
  for (long long int i = 1;; i++) {
    if (res >= c) {
      cout << i - 1 << endl;
      return 0;
    }
    res += cnt(i);
  }
}

```