#include <bits/stdc++.h>
using namespace std;
long long n, x, y, c;
long long MR(long long f, long long m) {
  return (m > f) ? max(m - f, 0ll) + max((m - f) * (m - f - 1), 0ll) : 0;
}
long long G(long long f, long long m) {
  return (m + 1 > f) ? max((m - f + 1) * (m - f + 2) / 2, 0ll) : 0;
}
long long t(long long m) {
  if (m == -1) return 0;
  long long T = (m > 0) ? (m * 4 + m * (m - 1) * 2 + 1) : 1;
  long long A = MR(y - 1, m) + MR(n - y, m) + MR(x - 1, m) + MR(n - x, m);
  long long K = G(x + y, m) + G(n - x + 1 + y, m) + G(x + n - y + 1, m) +
                G(n - x + n - y + 2, m);
  return T - A + K;
}
int main() {
  cin >> n >> x >> y >> c;
  long long k = 0, b = n, m = 0;
  while (t(m) < c) m++;
  while (!((t(m) >= c) && t(m - 1) < c)) {
    if (t(m) > c)
      b = m;
    else
      k = m;
    m = (b + k) / 2;
  }
  cout << m << endl;
  return 0;
}

### Prompt

Your task is to create 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 MR(long long f, long long m) {
  return (m > f) ? max(m - f, 0ll) + max((m - f) * (m - f - 1), 0ll) : 0;
}
long long G(long long f, long long m) {
  return (m + 1 > f) ? max((m - f + 1) * (m - f + 2) / 2, 0ll) : 0;
}
long long t(long long m) {
  if (m == -1) return 0;
  long long T = (m > 0) ? (m * 4 + m * (m - 1) * 2 + 1) : 1;
  long long A = MR(y - 1, m) + MR(n - y, m) + MR(x - 1, m) + MR(n - x, m);
  long long K = G(x + y, m) + G(n - x + 1 + y, m) + G(x + n - y + 1, m) +
                G(n - x + n - y + 2, m);
  return T - A + K;
}
int main() {
  cin >> n >> x >> y >> c;
  long long k = 0, b = n, m = 0;
  while (t(m) < c) m++;
  while (!((t(m) >= c) && t(m - 1) < c)) {
    if (t(m) > c)
      b = m;
    else
      k = m;
    m = (b + k) / 2;
  }
  cout << m << endl;
  return 0;
}

```