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

### Prompt

Please create a solution in cpp 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 cal(long long n, long long m, long long t) {
  if (n > m) swap(n, m);
  if (t <= n) return t * (t + 1) / 2;
  if (t <= m) return (t - n) * n + n * (n + 1) / 2;
  t = max(n + m - 1 - t, 0LL);
  return n * m - t * (t + 1) / 2;
}
int main() {
  int n, x, y, c, l, r, mid, ans;
  scanf("%d%d%d%d", &n, &x, &y, &c);
  --c;
  for (l = 0, r = 2 * n; l <= r;) {
    mid = l + (r - l) / 2;
    if (cal(x - 1, n - y + 1, mid) + cal(n - x + 1, n - y, mid) +
            cal(n - x, y, mid) + cal(x, y - 1, mid) <
        c)
      l = mid + 1;
    else
      ans = mid, r = mid - 1;
  }
  printf("%d", ans);
}

```