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

### Prompt

Generate 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, c;
long long int calc(long long int len) {
  long long int ans = len * len + (len + 1) * (len + 1);
  long long int t = x + len - n;
  if (t > 0) ans = ans - t * t;
  t = len - x + 1;
  if (t > 0) ans = ans - t * t;
  t = y + len - n;
  if (t > 0) ans = ans - t * t;
  t = len - y + 1;
  if (t > 0) ans = ans - t * t;
  t = -(x + (n + 1 - y) - len);
  t++;
  if (t > 0) ans = ans + (t + 1) * t / 2;
  t = y + (n + 1 - x) - len;
  t = -t;
  t++;
  if (t > 0) ans = ans + (t + 1) * t / 2;
  t = x + y - len;
  t = -t;
  t++;
  if (t > 0) ans = ans + (t + 1) * t / 2;
  t = n + 1 - x + n + 1 - y - len;
  t = -t;
  t++;
  if (t > 0) ans = ans + (t + 1) * t / 2;
  return ans;
}
int main() {
  cin >> n >> x >> y >> c;
  long long int l = 0, r = n + n;
  while (l < r) {
    long long int mid = (l + r) / 2;
    long long int s = calc(mid);
    if (s >= c)
      r = mid;
    else
      l = mid + 1;
  }
  cout << r << endl;
}

```