#include <bits/stdc++.h>
using namespace std;
long long n, x, y, c;
long long sq(long long x) { return (x * (x + 1)) / 2LL; }
long long g(long long x, long long y, long long t) {
  if (x <= 0 or x > n or y <= 0 or y > n or t < 0) return 0;
  long long ans = x * y;
  if (x + y <= 2LL + t) return ans;
  ans = sq(t + 1);
  if (x - t <= 0) ans -= sq(1 + t - x);
  if (y - t <= 0) ans -= sq(1 + t - y);
  return ans;
}
bool P(long long mid) {
  long long ans = 1;
  if (mid == 0) return ans >= c;
  ans = 0;
  ans += g(x, y, mid);
  ans += g(x, n - y, mid - 1);
  ans += g(n - x, y, mid - 1);
  ans += g(n - x, n - y, mid - 2);
  return ans >= c;
}
int main() {
  ios ::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  cin >> n >> x >> y >> c;
  long long l = 0;
  long long r = 2e9;
  while (l < r) {
    long long mid = (l + r) / 2;
    if (P(mid))
      r = mid;
    else
      l = mid + 1;
  }
  cout << l << "\n";
}

### Prompt

Construct a cpp code solution to the problem outlined:

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 sq(long long x) { return (x * (x + 1)) / 2LL; }
long long g(long long x, long long y, long long t) {
  if (x <= 0 or x > n or y <= 0 or y > n or t < 0) return 0;
  long long ans = x * y;
  if (x + y <= 2LL + t) return ans;
  ans = sq(t + 1);
  if (x - t <= 0) ans -= sq(1 + t - x);
  if (y - t <= 0) ans -= sq(1 + t - y);
  return ans;
}
bool P(long long mid) {
  long long ans = 1;
  if (mid == 0) return ans >= c;
  ans = 0;
  ans += g(x, y, mid);
  ans += g(x, n - y, mid - 1);
  ans += g(n - x, y, mid - 1);
  ans += g(n - x, n - y, mid - 2);
  return ans >= c;
}
int main() {
  ios ::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  cin >> n >> x >> y >> c;
  long long l = 0;
  long long r = 2e9;
  while (l < r) {
    long long mid = (l + r) / 2;
    if (P(mid))
      r = mid;
    else
      l = mid + 1;
  }
  cout << l << "\n";
}

```