#include <bits/stdc++.h>
using namespace std;
long long c, l, r, mid, n, x, y;
long long cal(long long x) { return (2 * x) * x >> 1; }
long long cal2(long long x) { return (x + 1) * x >> 1; }
bool check(long long t) {
  long long sss = (t + 1 + t) * (t + 1) - t;
  if (x + t > n) sss -= cal(x + t - n);
  if (x - t <= 0) sss -= cal(t - x + 1);
  if (y + t > n) sss -= cal(y + t - n);
  if (y - t <= 0) sss -= cal(t - y + 1);
  if (t - x - y + 1 > 0) sss += cal2(t - x - y + 1);
  if (x + n - y < t) sss += cal2(t - x - n + y);
  if (y + n - x < t) sss += cal2(t - y - n + x);
  if (n - x + n - y + 1 < t) sss += cal2(t - n - n + x + y - 1);
  return sss >= c;
}
int main() {
  scanf("%I64d%I64d%I64d%I64d", &n, &x, &y, &c);
  for (l = -1, r = 2000000000LL; l < r - 1;)
    if (check(mid = l + r >> 1))
      r = mid;
    else
      l = mid;
  printf("%I64d", r);
  return 0;
}

### Prompt

In Cpp, your task is to solve 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 c, l, r, mid, n, x, y;
long long cal(long long x) { return (2 * x) * x >> 1; }
long long cal2(long long x) { return (x + 1) * x >> 1; }
bool check(long long t) {
  long long sss = (t + 1 + t) * (t + 1) - t;
  if (x + t > n) sss -= cal(x + t - n);
  if (x - t <= 0) sss -= cal(t - x + 1);
  if (y + t > n) sss -= cal(y + t - n);
  if (y - t <= 0) sss -= cal(t - y + 1);
  if (t - x - y + 1 > 0) sss += cal2(t - x - y + 1);
  if (x + n - y < t) sss += cal2(t - x - n + y);
  if (y + n - x < t) sss += cal2(t - y - n + x);
  if (n - x + n - y + 1 < t) sss += cal2(t - n - n + x + y - 1);
  return sss >= c;
}
int main() {
  scanf("%I64d%I64d%I64d%I64d", &n, &x, &y, &c);
  for (l = -1, r = 2000000000LL; l < r - 1;)
    if (check(mid = l + r >> 1))
      r = mid;
    else
      l = mid;
  printf("%I64d", r);
  return 0;
}

```