#include <bits/stdc++.h>
using namespace std;
long long n, x, y, c;
long long g(long long k) { return k * k; }
long long h(long long k) { return ((k + 1) * (k + 2)) / 2; }
long long f(long long d) {
  long long s = d * d + (d + 1) * (d + 1);
  if (d - (y - 1) >= 1) s -= g(d - (y - 1));
  if (d - (n - y) >= 1) s -= g(d - (n - y));
  if (d - (x - 1) >= 1) s -= g(d - (x - 1));
  if (d - (n - x) >= 1) s -= g(d - (n - x));
  if (d - (y - 1) - (x - 1) >= 2) s += h(d - (y - 1) - (x - 1) - 2);
  if (d - (y - 1) - (n - x) >= 2) s += h(d - (y - 1) - (n - x) - 2);
  if (d - (n - y) - (x - 1) >= 2) s += h(d - (n - y) - (x - 1) - 2);
  if (d - (n - y) - (n - x) >= 2) s += h(d - (n - y) - (n - x) - 2);
  return s;
}
int main() {
  scanf("%I64d%I64d%I64d%I64d", &n, &x, &y, &c);
  long long ans = 2 * n, left = 0, right = 2 * n;
  while (left <= right) {
    long long mid = (left + right) >> 1;
    if (f(mid) >= c) {
      ans = mid;
      right = mid - 1;
    } else
      left = mid + 1;
  }
  printf("%I64d\n", ans);
  return 0;
}

### Prompt

Your task is to create 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 g(long long k) { return k * k; }
long long h(long long k) { return ((k + 1) * (k + 2)) / 2; }
long long f(long long d) {
  long long s = d * d + (d + 1) * (d + 1);
  if (d - (y - 1) >= 1) s -= g(d - (y - 1));
  if (d - (n - y) >= 1) s -= g(d - (n - y));
  if (d - (x - 1) >= 1) s -= g(d - (x - 1));
  if (d - (n - x) >= 1) s -= g(d - (n - x));
  if (d - (y - 1) - (x - 1) >= 2) s += h(d - (y - 1) - (x - 1) - 2);
  if (d - (y - 1) - (n - x) >= 2) s += h(d - (y - 1) - (n - x) - 2);
  if (d - (n - y) - (x - 1) >= 2) s += h(d - (n - y) - (x - 1) - 2);
  if (d - (n - y) - (n - x) >= 2) s += h(d - (n - y) - (n - x) - 2);
  return s;
}
int main() {
  scanf("%I64d%I64d%I64d%I64d", &n, &x, &y, &c);
  long long ans = 2 * n, left = 0, right = 2 * n;
  while (left <= right) {
    long long mid = (left + right) >> 1;
    if (f(mid) >= c) {
      ans = mid;
      right = mid - 1;
    } else
      left = mid + 1;
  }
  printf("%I64d\n", ans);
  return 0;
}

```