#include <bits/stdc++.h>
using namespace std;
long long n, x, y, c;
long long calc(long long k) {
  long long ans = (k + 1) * (k + 1) + k * k;
  if (k + 1 > x) ans -= (k + 1 - x) * (k + 1 - x);
  if (k >= n - x) ans -= (k - (n - x)) * (k - (n - x));
  if (k + 1 > y) {
    long long end = k + 1 - y;
    long long first = max(1ll, k + 1 - y - (x - 1));
    ans -= (end + first) * (end - first + 1) / 2;
  }
  if (k > n - y) {
    long long end = k - (n - y);
    long long first = max(1ll, k - (n - y) - (x - 1));
    ans -= (end + first) * (end - first + 1) / 2;
  }
  if (k > y) {
    long long end = k - y;
    long long first = max(1ll, k + 1 - y - (n - x));
    ans -= (end + first) * (end - first + 1) / 2;
  }
  if (k - 1 > n - y) {
    long long end = k - (n - y) - 1;
    long long first = max(1ll, k - (n - y) - (n - x));
    ans -= (end + first) * (end - first + 1) / 2;
  }
  return ans;
}
int main() {
  cin >> n >> y >> x >> c;
  long long f = 0;
  long long e = c + 1;
  long long bestAns = e;
  while (f < e) {
    long long mid = (f + e) / 2;
    if (calc(mid) >= c) {
      bestAns = mid;
      e = mid;
    } else
      f = mid + 1;
  }
  cout << bestAns << endl;
}

### Prompt

Your challenge is to write 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 calc(long long k) {
  long long ans = (k + 1) * (k + 1) + k * k;
  if (k + 1 > x) ans -= (k + 1 - x) * (k + 1 - x);
  if (k >= n - x) ans -= (k - (n - x)) * (k - (n - x));
  if (k + 1 > y) {
    long long end = k + 1 - y;
    long long first = max(1ll, k + 1 - y - (x - 1));
    ans -= (end + first) * (end - first + 1) / 2;
  }
  if (k > n - y) {
    long long end = k - (n - y);
    long long first = max(1ll, k - (n - y) - (x - 1));
    ans -= (end + first) * (end - first + 1) / 2;
  }
  if (k > y) {
    long long end = k - y;
    long long first = max(1ll, k + 1 - y - (n - x));
    ans -= (end + first) * (end - first + 1) / 2;
  }
  if (k - 1 > n - y) {
    long long end = k - (n - y) - 1;
    long long first = max(1ll, k - (n - y) - (n - x));
    ans -= (end + first) * (end - first + 1) / 2;
  }
  return ans;
}
int main() {
  cin >> n >> y >> x >> c;
  long long f = 0;
  long long e = c + 1;
  long long bestAns = e;
  while (f < e) {
    long long mid = (f + e) / 2;
    if (calc(mid) >= c) {
      bestAns = mid;
      e = mid;
    } else
      f = mid + 1;
  }
  cout << bestAns << endl;
}

```