#include <bits/stdc++.h>
using namespace std;
int n, c, cr, cc;
long long a[3000000];
inline long long triangle(long long l) {
  if (l < 0) return 0;
  return l * l;
}
inline long long corner(long long l) {
  if (l < 0) return 0;
  return (l * (l + 1)) / 2;
}
inline long long kill(int t) {
  long long res = 0;
  if (!t) return 0;
  res += triangle(t - cc + 1);
  res += triangle(t - cr + 1);
  res += triangle(t - n + cc);
  res += triangle(t - n + cr);
  res -= corner(t - cr - cc + 1);
  res -= corner(t - cr - n + cc);
  res -= corner(t - n + cr - cc);
  res -= corner(t + cc + cr - 2 * n - 1);
  return res;
}
int main() {
  cin >> n >> cr >> cc >> c;
  int cnt;
  a[0] = 1;
  for (cnt = 1; a[cnt - 1] < 10000000000000ll; ++cnt)
    a[cnt] = a[cnt - 1] + cnt * 4;
  int t = 0;
  while ((a[t] - kill(t)) < c) {
    ++t;
  }
  cout << t;
  return 0;
}

### 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;
int n, c, cr, cc;
long long a[3000000];
inline long long triangle(long long l) {
  if (l < 0) return 0;
  return l * l;
}
inline long long corner(long long l) {
  if (l < 0) return 0;
  return (l * (l + 1)) / 2;
}
inline long long kill(int t) {
  long long res = 0;
  if (!t) return 0;
  res += triangle(t - cc + 1);
  res += triangle(t - cr + 1);
  res += triangle(t - n + cc);
  res += triangle(t - n + cr);
  res -= corner(t - cr - cc + 1);
  res -= corner(t - cr - n + cc);
  res -= corner(t - n + cr - cc);
  res -= corner(t + cc + cr - 2 * n - 1);
  return res;
}
int main() {
  cin >> n >> cr >> cc >> c;
  int cnt;
  a[0] = 1;
  for (cnt = 1; a[cnt - 1] < 10000000000000ll; ++cnt)
    a[cnt] = a[cnt - 1] + cnt * 4;
  int t = 0;
  while ((a[t] - kill(t)) < c) {
    ++t;
  }
  cout << t;
  return 0;
}

```