#include <bits/stdc++.h>
using namespace std;
long long x, y, n, c;
long long nhan(long long u) {
  u = max(u, 0ll);
  return u * (u + 1) / 2;
}
bool kt(long long t) {
  long long p = 0;
  if (t < 2 * n - x - y)
    p += nhan(t + 1) - nhan(y + t - n) - nhan(x + t - n);
  else
    p += (n - x + 1) * (n - y + 1);
  if (t < n - x + y - 1)
    p += nhan(t + 1) - nhan(t - y + 1) - nhan(x + t - n);
  else
    p += (n - x + 1) * (y);
  if (t < x - 1 + y - 1)
    p += nhan(t + 1) - nhan(t - y + 1) - nhan(t - x + 1);
  else
    p += x * y;
  if (t < x - 1 + n - y)
    p += nhan(t + 1) - nhan(t - x + 1) - nhan(y + t - n);
  else
    p += (x) * (n - y + 1);
  p -= min(t, x - 1) + min(t, n - x) + min(t, n - y) + min(t, y - 1) + 3;
  return p >= c;
}
int main() {
  cin >> n >> x >> y >> c;
  if (n == 1) {
    cout << 0;
    return 0;
  }
  long long l = -1;
  long long r = 1ll * 1e13;
  while (r - l > 1) {
    long long t = (l + r) / 2;
    if (kt(t))
      r = t;
    else
      l = t;
  }
  cout << r;
}

### 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 x, y, n, c;
long long nhan(long long u) {
  u = max(u, 0ll);
  return u * (u + 1) / 2;
}
bool kt(long long t) {
  long long p = 0;
  if (t < 2 * n - x - y)
    p += nhan(t + 1) - nhan(y + t - n) - nhan(x + t - n);
  else
    p += (n - x + 1) * (n - y + 1);
  if (t < n - x + y - 1)
    p += nhan(t + 1) - nhan(t - y + 1) - nhan(x + t - n);
  else
    p += (n - x + 1) * (y);
  if (t < x - 1 + y - 1)
    p += nhan(t + 1) - nhan(t - y + 1) - nhan(t - x + 1);
  else
    p += x * y;
  if (t < x - 1 + n - y)
    p += nhan(t + 1) - nhan(t - x + 1) - nhan(y + t - n);
  else
    p += (x) * (n - y + 1);
  p -= min(t, x - 1) + min(t, n - x) + min(t, n - y) + min(t, y - 1) + 3;
  return p >= c;
}
int main() {
  cin >> n >> x >> y >> c;
  if (n == 1) {
    cout << 0;
    return 0;
  }
  long long l = -1;
  long long r = 1ll * 1e13;
  while (r - l > 1) {
    long long t = (l + r) / 2;
    if (kt(t))
      r = t;
    else
      l = t;
  }
  cout << r;
}

```