#include <bits/stdc++.h>
using namespace std;
int n, a, b;
long long cal(int x) {
  long long res = 1 + 2 * (long long)x * (x + 1);
  long long tmp;
  tmp = x + 1 - a - b;
  if (tmp > 0) res += tmp * (tmp + 1) / 2;
  tmp = x + a - b - n;
  if (tmp > 0) res += tmp * (tmp + 1) / 2;
  tmp = x + b - a - n;
  if (tmp > 0) res += tmp * (tmp + 1) / 2;
  tmp = x + a + b - 2 * n - 1;
  if (tmp > 0) res += tmp * (tmp + 1) / 2;
  tmp = x + 1 - a;
  if (tmp > 0) res -= tmp * tmp;
  tmp = x + 1 - b;
  if (tmp > 0) res -= tmp * tmp;
  tmp = x + a - n;
  if (tmp > 0) res -= tmp * tmp;
  tmp = x + b - n;
  if (tmp > 0) res -= tmp * tmp;
  return res;
}
int main() {
  long long c;
  int l, r, x;
  scanf("%d%d%d%I64d", &n, &a, &b, &c);
  l = 0;
  r = 2 * n;
  while (l != r) {
    x = l + r >> 1;
    if (cal(x) < c)
      l = x + 1;
    else
      r = x;
  }
  printf("%d\n", l);
}

### 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;
int n, a, b;
long long cal(int x) {
  long long res = 1 + 2 * (long long)x * (x + 1);
  long long tmp;
  tmp = x + 1 - a - b;
  if (tmp > 0) res += tmp * (tmp + 1) / 2;
  tmp = x + a - b - n;
  if (tmp > 0) res += tmp * (tmp + 1) / 2;
  tmp = x + b - a - n;
  if (tmp > 0) res += tmp * (tmp + 1) / 2;
  tmp = x + a + b - 2 * n - 1;
  if (tmp > 0) res += tmp * (tmp + 1) / 2;
  tmp = x + 1 - a;
  if (tmp > 0) res -= tmp * tmp;
  tmp = x + 1 - b;
  if (tmp > 0) res -= tmp * tmp;
  tmp = x + a - n;
  if (tmp > 0) res -= tmp * tmp;
  tmp = x + b - n;
  if (tmp > 0) res -= tmp * tmp;
  return res;
}
int main() {
  long long c;
  int l, r, x;
  scanf("%d%d%d%I64d", &n, &a, &b, &c);
  l = 0;
  r = 2 * n;
  while (l != r) {
    x = l + r >> 1;
    if (cal(x) < c)
      l = x + 1;
    else
      r = x;
  }
  printf("%d\n", l);
}

```