#include <bits/stdc++.h>
using namespace std;
inline long long tra(long long x) {
  if (x <= 0) return 0;
  return x * (x + 1) / 2;
}
inline long long tra2(long long x) {
  if (x <= 0) return 0;
  return x * x;
  ;
}
long long N, c, x, y;
int main(void) {
  std::ios_base::sync_with_stdio(false);
  cin >> N >> x >> y >> c;
  long long da = 0, a = INT_MAX;
  while (1) {
    long long t = (da + a) / 2;
    if (da == a) {
      cout << da;
      return 0;
    }
    long long dopox2 = tra2(t - (N + 1 - x) + 1);
    long long primax1 = tra2(t - (x) + 1);
    long long primay2 = tra2(t - (N + 1 - y) + 1);
    long long dopoy1 = tra2(t - (y) + 1);
    long long dopox2y1 = tra(t - (y + N + 1 - x) + 1);
    long long dopox2y2 = tra(t - (N + 1 - y + N + 1 - x) + 1);
    long long dopox1y1 = tra(t - (x + y) + 1);
    long long dopox1y2 = tra(t - (x + N + 1 - y) + 1);
    long long num = 2 * t * t + 2 * t + 1 - primax1 - primay2 - dopoy1 -
                    dopox2 + dopox2y1 + dopox2y2 + dopox1y2 + dopox1y1;
    if (num >= c)
      a = t;
    else
      da = t + 1;
  }
  return 0;
}

### Prompt

Please formulate 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;
inline long long tra(long long x) {
  if (x <= 0) return 0;
  return x * (x + 1) / 2;
}
inline long long tra2(long long x) {
  if (x <= 0) return 0;
  return x * x;
  ;
}
long long N, c, x, y;
int main(void) {
  std::ios_base::sync_with_stdio(false);
  cin >> N >> x >> y >> c;
  long long da = 0, a = INT_MAX;
  while (1) {
    long long t = (da + a) / 2;
    if (da == a) {
      cout << da;
      return 0;
    }
    long long dopox2 = tra2(t - (N + 1 - x) + 1);
    long long primax1 = tra2(t - (x) + 1);
    long long primay2 = tra2(t - (N + 1 - y) + 1);
    long long dopoy1 = tra2(t - (y) + 1);
    long long dopox2y1 = tra(t - (y + N + 1 - x) + 1);
    long long dopox2y2 = tra(t - (N + 1 - y + N + 1 - x) + 1);
    long long dopox1y1 = tra(t - (x + y) + 1);
    long long dopox1y2 = tra(t - (x + N + 1 - y) + 1);
    long long num = 2 * t * t + 2 * t + 1 - primax1 - primay2 - dopoy1 -
                    dopox2 + dopox2y1 + dopox2y2 + dopox1y2 + dopox1y1;
    if (num >= c)
      a = t;
    else
      da = t + 1;
  }
  return 0;
}

```