#include <bits/stdc++.h>
using namespace std;
long long n, x, y, c;
long long levo, desno, gor, dol;
long long kar(long long k) { return 1 + 2 * k * (k + 1); }
long long tri(long long k) { return k * k; }
long long vog(long long k) { return k * (k + 1) / 2; }
long long on(int k) {
  long long o = kar(k);
  if (k > desno) o -= tri(k - desno);
  if (k > levo) o -= tri(k - levo);
  if (k > gor) o -= tri(k - gor);
  if (k > dol) o -= tri(k - dol);
  if (k - desno - gor - 1 > 0) o += vog(k - desno - gor - 1);
  if (k - desno - dol - 1 > 0) o += vog(k - desno - dol - 1);
  if (k - levo - gor - 1 > 0) o += vog(k - levo - gor - 1);
  if (k - levo - dol - 1 > 0) o += vog(k - levo - dol - 1);
  return o;
}
int main() {
  cin >> n >> x >> y >> c;
  levo = x - 1;
  desno = n - x;
  gor = y - 1;
  dol = n - y;
  if (c == 1) {
    cout << 0 << endl;
    return 0;
  }
  int m = 0, M = 1000000000;
  while (M - m > 1) {
    int k = (M + m) / 2;
    if (on(k) >= c)
      M = k;
    else
      m = k;
  }
  cout << M << endl;
  return 0;
}

### 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 levo, desno, gor, dol;
long long kar(long long k) { return 1 + 2 * k * (k + 1); }
long long tri(long long k) { return k * k; }
long long vog(long long k) { return k * (k + 1) / 2; }
long long on(int k) {
  long long o = kar(k);
  if (k > desno) o -= tri(k - desno);
  if (k > levo) o -= tri(k - levo);
  if (k > gor) o -= tri(k - gor);
  if (k > dol) o -= tri(k - dol);
  if (k - desno - gor - 1 > 0) o += vog(k - desno - gor - 1);
  if (k - desno - dol - 1 > 0) o += vog(k - desno - dol - 1);
  if (k - levo - gor - 1 > 0) o += vog(k - levo - gor - 1);
  if (k - levo - dol - 1 > 0) o += vog(k - levo - dol - 1);
  return o;
}
int main() {
  cin >> n >> x >> y >> c;
  levo = x - 1;
  desno = n - x;
  gor = y - 1;
  dol = n - y;
  if (c == 1) {
    cout << 0 << endl;
    return 0;
  }
  int m = 0, M = 1000000000;
  while (M - m > 1) {
    int k = (M + m) / 2;
    if (on(k) >= c)
      M = k;
    else
      m = k;
  }
  cout << M << endl;
  return 0;
}

```