#include <bits/stdc++.h>
using namespace std;
long long f(long long x, long long y, long long n, long long c) {
  if (c == -1) return 0LL;
  long long ret = (c + 1) * (c + 1) + c * c;
  if (c > x - 1) ret -= (c - x + 1) * (c - x + 1);
  if (c > n - x) ret -= (c + x - n) * (c + x - n);
  if (c > y - 1) ret -= (c - y + 1) * (c - y + 1);
  if (c > n - y + 1) ret -= (c + y - n) * (c + y - n);
  if (c > x + y - 1) ret += (c - x - y + 1) * (c - x - y + 2) / 2;
  if (c > x + n - y) ret += (c + y - x - n) * (c + y - x - n + 1) / 2;
  if (c > n - x + y) ret += (c + x - y - n) * (c + x - y - n + 1) / 2;
  if (c > 2 * n - x - y + 1)
    ret += (c + x + y - 2 * n) * (c + x + y - 2 * n - 1) / 2;
  return ret;
}
long long pesquisa(long long x, long long y, long long n, long long c) {
  long long ini = 0LL, fim = (long long)1000000000, meio = (ini + fim) >> 1;
  for (; ini <= fim; meio = (ini + fim) >> 1) {
    if (f(x, y, n, meio) >= c && f(x, y, n, meio - 1) < c)
      return meio;
    else if (f(x, y, n, meio) < c)
      ini = meio + 1;
    else
      fim = meio - 1;
  }
}
int main() {
  long long n, x, y, c;
  cin >> n >> x >> y >> c;
  cout << pesquisa(x, y, n, c) << endl;
  return 0;
}

### 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 f(long long x, long long y, long long n, long long c) {
  if (c == -1) return 0LL;
  long long ret = (c + 1) * (c + 1) + c * c;
  if (c > x - 1) ret -= (c - x + 1) * (c - x + 1);
  if (c > n - x) ret -= (c + x - n) * (c + x - n);
  if (c > y - 1) ret -= (c - y + 1) * (c - y + 1);
  if (c > n - y + 1) ret -= (c + y - n) * (c + y - n);
  if (c > x + y - 1) ret += (c - x - y + 1) * (c - x - y + 2) / 2;
  if (c > x + n - y) ret += (c + y - x - n) * (c + y - x - n + 1) / 2;
  if (c > n - x + y) ret += (c + x - y - n) * (c + x - y - n + 1) / 2;
  if (c > 2 * n - x - y + 1)
    ret += (c + x + y - 2 * n) * (c + x + y - 2 * n - 1) / 2;
  return ret;
}
long long pesquisa(long long x, long long y, long long n, long long c) {
  long long ini = 0LL, fim = (long long)1000000000, meio = (ini + fim) >> 1;
  for (; ini <= fim; meio = (ini + fim) >> 1) {
    if (f(x, y, n, meio) >= c && f(x, y, n, meio - 1) < c)
      return meio;
    else if (f(x, y, n, meio) < c)
      ini = meio + 1;
    else
      fim = meio - 1;
  }
}
int main() {
  long long n, x, y, c;
  cin >> n >> x >> y >> c;
  cout << pesquisa(x, y, n, c) << endl;
  return 0;
}

```