#include <bits/stdc++.h>
using namespace std;
long long n, x, y, c;
long long calc1(long long len, long long n, long long m) {
  len++;
  if (len <= n && len <= m) return len * (len + 1) / 2;
  if (len <= n) return (len + len - m + 1) * m / 2;
  if (len <= m) return (len + len - n + 1) * n / 2;
  len = max(n + m - len - 1, 0LL);
  return n * m - len * (len + 1) / 2;
}
long long calc(long long len) {
  if (len == 0) return 1;
  if (len == 1) return calc1(len, x, y) + calc1(len, n - x + 1, n - y + 1) - 1;
  return calc1(len, x, y) + calc1(len, n - x + 1, n - y + 1) +
         calc1(len - 2, n - x, y - 1) + calc1(len - 2, x - 1, n - y) - 1;
}
int main() {
  while (cin >> n >> x >> y >> c) {
    long long l = 0, r = n * 2;
    while (l < r) {
      long long m = (l + r) / 2;
      if (calc(m) >= c)
        r = m;
      else
        l = m + 1;
    }
    cout << l << endl;
  }
  return 0;
}

### 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 n, x, y, c;
long long calc1(long long len, long long n, long long m) {
  len++;
  if (len <= n && len <= m) return len * (len + 1) / 2;
  if (len <= n) return (len + len - m + 1) * m / 2;
  if (len <= m) return (len + len - n + 1) * n / 2;
  len = max(n + m - len - 1, 0LL);
  return n * m - len * (len + 1) / 2;
}
long long calc(long long len) {
  if (len == 0) return 1;
  if (len == 1) return calc1(len, x, y) + calc1(len, n - x + 1, n - y + 1) - 1;
  return calc1(len, x, y) + calc1(len, n - x + 1, n - y + 1) +
         calc1(len - 2, n - x, y - 1) + calc1(len - 2, x - 1, n - y) - 1;
}
int main() {
  while (cin >> n >> x >> y >> c) {
    long long l = 0, r = n * 2;
    while (l < r) {
      long long m = (l + r) / 2;
      if (calc(m) >= c)
        r = m;
      else
        l = m + 1;
    }
    cout << l << endl;
  }
  return 0;
}

```