#include <bits/stdc++.h>
using namespace std;
long long count(long long n, long long x, long long y, long long turn) {
  if (!turn) return 1;
  long long res = 0;
  long long xl = x - 1, xr = n - x, y0 = min(turn, y - 1), yy;
  yy = max(1LL, turn - xl);
  yy = min(yy, y0 + 1);
  res += (turn - y0 + turn - yy) * (y0 - yy + 1) / 2;
  res += (yy - 1) * xl;
  yy = max(1LL, turn - xr);
  yy = min(yy, y0 + 1);
  res += (turn - y0 + turn - yy) * (y0 - yy + 1) / 2;
  res += (yy - 1) * xr;
  y0 = min(turn, n - y);
  yy = max(1LL, turn - xl);
  yy = min(yy, y0 + 1);
  res += (turn - y0 + turn - yy) * (y0 - yy + 1) / 2;
  res += (yy - 1) * xl;
  yy = max(1LL, turn - xr);
  yy = min(yy, y0 + 1);
  res += (turn - y0 + turn - yy) * (y0 - yy + 1) / 2;
  res += (yy - 1) * xr;
  res += min(n, y + turn) - max(1LL, y - turn) + min(n, x + turn) -
         max(1LL, x - turn) + 1;
  return res;
}
int main() {
  long long n, x, y, c;
  cin >> n >> x >> y >> c;
  long long low = 0, high = n * 3, ans;
  while (low <= high) {
    long long mid = (low + high) / 2;
    if (count(n, x, y, mid) >= c)
      ans = mid, high = mid - 1;
    else
      low = mid + 1;
  }
  cout << ans << endl;
}

### Prompt

Develop a solution in CPP 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 count(long long n, long long x, long long y, long long turn) {
  if (!turn) return 1;
  long long res = 0;
  long long xl = x - 1, xr = n - x, y0 = min(turn, y - 1), yy;
  yy = max(1LL, turn - xl);
  yy = min(yy, y0 + 1);
  res += (turn - y0 + turn - yy) * (y0 - yy + 1) / 2;
  res += (yy - 1) * xl;
  yy = max(1LL, turn - xr);
  yy = min(yy, y0 + 1);
  res += (turn - y0 + turn - yy) * (y0 - yy + 1) / 2;
  res += (yy - 1) * xr;
  y0 = min(turn, n - y);
  yy = max(1LL, turn - xl);
  yy = min(yy, y0 + 1);
  res += (turn - y0 + turn - yy) * (y0 - yy + 1) / 2;
  res += (yy - 1) * xl;
  yy = max(1LL, turn - xr);
  yy = min(yy, y0 + 1);
  res += (turn - y0 + turn - yy) * (y0 - yy + 1) / 2;
  res += (yy - 1) * xr;
  res += min(n, y + turn) - max(1LL, y - turn) + min(n, x + turn) -
         max(1LL, x - turn) + 1;
  return res;
}
int main() {
  long long n, x, y, c;
  cin >> n >> x >> y >> c;
  long long low = 0, high = n * 3, ans;
  while (low <= high) {
    long long mid = (low + high) / 2;
    if (count(n, x, y, mid) >= c)
      ans = mid, high = mid - 1;
    else
      low = mid + 1;
  }
  cout << ans << endl;
}

```