#include <bits/stdc++.h>
using namespace std;
long long n, x, y, c;
long long cal(long long x, long long y, long long z) {
  if (x > y) swap(x, y);
  if (!x) return 0;
  if (z <= x) return z * (z + 1) / 2;
  if (z <= y) return (z - x) * x + x * (x + 1) / 2;
  z = max(0ll, x + y - 1 - z);
  return x * y - z * (z + 1) / 2;
}
bool chk(long long now) {
  long long t = 0;
  t += cal(x - 1, y - 1, now - 1);
  t += cal(n - x + 1, y, now + 1);
  t += cal(x, n - y + 1, now + 1);
  t += cal(n - x, n - y, now - 1);
  return t > c;
}
int main() {
  ios::sync_with_stdio(0);
  cin >> n >> x >> y >> c;
  long long l = 0, r = n * 2, mid, ans = 0;
  for (; l <= r;) chk(mid = l + r >> 1) ? r = (ans = mid) - 1 : l = mid + 1;
  cout << ans << endl;
}

### Prompt

Generate 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 cal(long long x, long long y, long long z) {
  if (x > y) swap(x, y);
  if (!x) return 0;
  if (z <= x) return z * (z + 1) / 2;
  if (z <= y) return (z - x) * x + x * (x + 1) / 2;
  z = max(0ll, x + y - 1 - z);
  return x * y - z * (z + 1) / 2;
}
bool chk(long long now) {
  long long t = 0;
  t += cal(x - 1, y - 1, now - 1);
  t += cal(n - x + 1, y, now + 1);
  t += cal(x, n - y + 1, now + 1);
  t += cal(n - x, n - y, now - 1);
  return t > c;
}
int main() {
  ios::sync_with_stdio(0);
  cin >> n >> x >> y >> c;
  long long l = 0, r = n * 2, mid, ans = 0;
  for (; l <= r;) chk(mid = l + r >> 1) ? r = (ans = mid) - 1 : l = mid + 1;
  cout << ans << endl;
}

```