#include <bits/stdc++.h>
using namespace std;
long long n, c, x, y, t;
long long in(long long t) {
  long long cnt = 0;
  if (x + y <= t)
    cnt += x * y;
  else {
    cnt += t * (t - 1) / 2;
    if (x < t) cnt -= (t - x) * (t - x - 1) / 2;
    if (y < t) cnt -= (t - y) * (t - y - 1) / 2;
  }
  return cnt;
}
long long space(long long t) {
  long long cnt = 1;
  cnt += min(x, t) + min(y, t);
  cnt += in(t);
  x = n - x - 1;
  cnt += in(t);
  y = n - y - 1;
  cnt += min(x, t) + min(y, t);
  cnt += in(t);
  x = n - x - 1;
  cnt += in(t);
  y = n - y - 1;
  return cnt;
}
int main() {
  cin >> n >> x >> y >> c;
  x--;
  y--;
  long long l = -1, r = n * 3, m;
  for (; r - l > 1;) {
    m = (r + l) >> 1;
    if (space(m) >= c)
      r = m;
    else
      l = m;
  }
  cout << r << endl;
  return 0;
}

### Prompt

Please provide a cpp coded solution to the problem described below:

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, c, x, y, t;
long long in(long long t) {
  long long cnt = 0;
  if (x + y <= t)
    cnt += x * y;
  else {
    cnt += t * (t - 1) / 2;
    if (x < t) cnt -= (t - x) * (t - x - 1) / 2;
    if (y < t) cnt -= (t - y) * (t - y - 1) / 2;
  }
  return cnt;
}
long long space(long long t) {
  long long cnt = 1;
  cnt += min(x, t) + min(y, t);
  cnt += in(t);
  x = n - x - 1;
  cnt += in(t);
  y = n - y - 1;
  cnt += min(x, t) + min(y, t);
  cnt += in(t);
  x = n - x - 1;
  cnt += in(t);
  y = n - y - 1;
  return cnt;
}
int main() {
  cin >> n >> x >> y >> c;
  x--;
  y--;
  long long l = -1, r = n * 3, m;
  for (; r - l > 1;) {
    m = (r + l) >> 1;
    if (space(m) >= c)
      r = m;
    else
      l = m;
  }
  cout << r << endl;
  return 0;
}

```