#include <bits/stdc++.h>
using namespace std;
long long x, y, n, c, l, r, mid;
long long calc1(long long a) {
  if (a < 0) return 0;
  return a * (a + 1) - a;
}
long long calc2(long long a) {
  if (a < 0) return 0;
  return a * (a + 1) / 2;
}
long long calc(long long d) {
  long long res = 1 + 4 / 2 * d * (d + 1);
  res -= calc1(d - (x - 1));
  res -= calc1(d - (y - 1));
  res -= calc1(d - (n - x));
  res -= calc1(d - (n - y));
  res += calc2(d - (x - 1) - (y - 1) - 1);
  res += calc2(d - (x - 1) - (n - y) - 1);
  res += calc2(d - (n - x) - (y - 1) - 1);
  res += calc2(d - (n - x) - (n - y) - 1);
  return res;
}
int main() {
  cin >> n >> x >> y >> c;
  l = -1;
  r = 2 * n + 10;
  while (l < r - 1) {
    mid = (l + r) / 2;
    if (calc(mid) < c)
      l = mid;
    else
      r = mid;
  }
  cout << r << "\n";
}

### Prompt

In Cpp, your task is to solve 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 x, y, n, c, l, r, mid;
long long calc1(long long a) {
  if (a < 0) return 0;
  return a * (a + 1) - a;
}
long long calc2(long long a) {
  if (a < 0) return 0;
  return a * (a + 1) / 2;
}
long long calc(long long d) {
  long long res = 1 + 4 / 2 * d * (d + 1);
  res -= calc1(d - (x - 1));
  res -= calc1(d - (y - 1));
  res -= calc1(d - (n - x));
  res -= calc1(d - (n - y));
  res += calc2(d - (x - 1) - (y - 1) - 1);
  res += calc2(d - (x - 1) - (n - y) - 1);
  res += calc2(d - (n - x) - (y - 1) - 1);
  res += calc2(d - (n - x) - (n - y) - 1);
  return res;
}
int main() {
  cin >> n >> x >> y >> c;
  l = -1;
  r = 2 * n + 10;
  while (l < r - 1) {
    mid = (l + r) / 2;
    if (calc(mid) < c)
      l = mid;
    else
      r = mid;
  }
  cout << r << "\n";
}

```