#include <bits/stdc++.h>
using namespace std;
long long triang(long long x) { return x * (x + 1) / 2; }
long long cant(long long x, long long y, long long t) {
  if (x < y) swap(x, y);
  long long k = max(0ll, t - x);
  if (k >= y) return x * y;
  return triang(y) - triang(k) + (x - (y - k)) * y;
}
long long solve(long long n, long long x, long long y, long long c) {
  if (c == 1) return 0;
  long long ini = 0, fin = 2 * n + 5, mid;
  while (ini < fin) {
    mid = (ini + fin) / 2;
    long long aux = 0, xx, yy;
    yy = min(mid, y), xx = min(mid, x);
    aux += cant(xx, yy, mid);
    aux -= yy + xx;
    yy = min(mid, y), xx = min(n - x + 1, mid);
    aux += cant(xx, yy, mid);
    yy = min(mid, n - y + 1), xx = min(mid, x);
    aux += cant(xx, yy, mid);
    yy = min(mid, n - y + 1), xx = min(mid, n - x + 1);
    aux += cant(xx, yy, mid);
    aux -= yy + xx;
    aux++;
    if (aux < c)
      ini = mid + 1;
    else
      fin = mid;
  }
  return ini - 1;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  long long n, x, y, c;
  cin >> n >> x >> y >> c;
  cout << solve(n, x, y, c) << "\n";
}

### Prompt

Your challenge is to write 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 triang(long long x) { return x * (x + 1) / 2; }
long long cant(long long x, long long y, long long t) {
  if (x < y) swap(x, y);
  long long k = max(0ll, t - x);
  if (k >= y) return x * y;
  return triang(y) - triang(k) + (x - (y - k)) * y;
}
long long solve(long long n, long long x, long long y, long long c) {
  if (c == 1) return 0;
  long long ini = 0, fin = 2 * n + 5, mid;
  while (ini < fin) {
    mid = (ini + fin) / 2;
    long long aux = 0, xx, yy;
    yy = min(mid, y), xx = min(mid, x);
    aux += cant(xx, yy, mid);
    aux -= yy + xx;
    yy = min(mid, y), xx = min(n - x + 1, mid);
    aux += cant(xx, yy, mid);
    yy = min(mid, n - y + 1), xx = min(mid, x);
    aux += cant(xx, yy, mid);
    yy = min(mid, n - y + 1), xx = min(mid, n - x + 1);
    aux += cant(xx, yy, mid);
    aux -= yy + xx;
    aux++;
    if (aux < c)
      ini = mid + 1;
    else
      fin = mid;
  }
  return ini - 1;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  long long n, x, y, c;
  cin >> n >> x >> y >> c;
  cout << solve(n, x, y, c) << "\n";
}

```