#include <bits/stdc++.h>
using namespace std;
long long n, x, y, c;
int main() {
  scanf("%I64d%I64d%I64d%I64d", &n, &x, &y, &c);
  x--;
  y--;
  long long k = 1, sec = 0;
  ;
  long long e_up = y, e_down = y, e_left = x, e_right = x;
  while (k < c) {
    sec++;
    k = ((long long)2 * sec + 1) * ((long long)2 * sec + 1) -
        (long long)2 * sec - 2 * sec * sec;
    e_up--;
    e_down++;
    e_left--;
    e_right++;
    if (sec == 6) {
      x = x + 1;
      x = x - 1;
    }
    if (e_up < 0) {
      long long d = -e_up;
      k -= (2 + (d - 1) * 2) / 2 * d;
    }
    if (e_left < 0) {
      long long d = -e_left;
      k -= (2 + (d - 1) * 2) / 2 * d;
    }
    if (e_down >= n) {
      long long d = e_down - n + 1;
      k -= (2 + (d - 1) * 2) / 2 * d;
    }
    if (e_right >= n) {
      long long d = e_right - n + 1;
      k -= (2 + (d - 1) * 2) / 2 * d;
    }
    if (x + 1 + y + 1 <= sec) {
      long long d = sec + 1 - (x + 1 + y + 1);
      k += (long long)(1 + d) * d / 2;
    }
    if (n - x + n - y <= sec) {
      long long d = sec + 1 - (n - x + n - y);
      k += (long long)(1 + d) * d / 2;
    }
    if (n - x + 1 + y <= sec) {
      long long d = sec + 1 - (n - x + y + 1);
      k += (long long)(1 + d) * d / 2;
    }
    if (x + 1 + n - y <= sec) {
      long long d = sec + 1 - (x + 1 + n - y);
      k += (long long)(1 + d) * d / 2;
    }
  }
  printf("%I64d", sec);
  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;
long long n, x, y, c;
int main() {
  scanf("%I64d%I64d%I64d%I64d", &n, &x, &y, &c);
  x--;
  y--;
  long long k = 1, sec = 0;
  ;
  long long e_up = y, e_down = y, e_left = x, e_right = x;
  while (k < c) {
    sec++;
    k = ((long long)2 * sec + 1) * ((long long)2 * sec + 1) -
        (long long)2 * sec - 2 * sec * sec;
    e_up--;
    e_down++;
    e_left--;
    e_right++;
    if (sec == 6) {
      x = x + 1;
      x = x - 1;
    }
    if (e_up < 0) {
      long long d = -e_up;
      k -= (2 + (d - 1) * 2) / 2 * d;
    }
    if (e_left < 0) {
      long long d = -e_left;
      k -= (2 + (d - 1) * 2) / 2 * d;
    }
    if (e_down >= n) {
      long long d = e_down - n + 1;
      k -= (2 + (d - 1) * 2) / 2 * d;
    }
    if (e_right >= n) {
      long long d = e_right - n + 1;
      k -= (2 + (d - 1) * 2) / 2 * d;
    }
    if (x + 1 + y + 1 <= sec) {
      long long d = sec + 1 - (x + 1 + y + 1);
      k += (long long)(1 + d) * d / 2;
    }
    if (n - x + n - y <= sec) {
      long long d = sec + 1 - (n - x + n - y);
      k += (long long)(1 + d) * d / 2;
    }
    if (n - x + 1 + y <= sec) {
      long long d = sec + 1 - (n - x + y + 1);
      k += (long long)(1 + d) * d / 2;
    }
    if (x + 1 + n - y <= sec) {
      long long d = sec + 1 - (x + 1 + n - y);
      k += (long long)(1 + d) * d / 2;
    }
  }
  printf("%I64d", sec);
  return 0;
}

```