#include <bits/stdc++.h>
using namespace std;
long long foo(long long x) { return ((x + 1) * x) / 2; }
int main() {
  long long n, x, y, c;
  while (scanf("%I64d%I64d%I64d%I64d", &n, &x, &y, &c) == 4) {
    long long es = 0, di = 2 * n;
    while (di > es) {
      long long me = (es + di) / 2;
      long long res = (me + 1) * (me + 1) + me * me;
      if (me - (n - x) > 0) res -= (me - (n - x)) * (me - (n - x));
      if (me - x + 1 > 0) res -= (me - x + 1) * (me - x + 1);
      if (me - (n - y) > 0) res -= (me - (n - y)) * (me - (n - y));
      if (me - y + 1 > 0) res -= (me - y + 1) * (me - y + 1);
      if (me - (n - x) - y > 0) res += foo(me - (n - x) - y);
      if (me - (n - x) - (n - y) - 1 > 0)
        res += foo(me - (n - x) - (n - y) - 1);
      if (me - x + 1 - y > 0) res += foo(me - x + 1 - y);
      if (me - x - (n - y) > 0) res += foo(me - x - (n - y));
      if (res >= c)
        di = me;
      else
        es = me + 1;
    }
    printf("%I64d\n", es);
  }
  return 0;
}

### Prompt

Construct a Cpp code solution to the problem outlined:

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 foo(long long x) { return ((x + 1) * x) / 2; }
int main() {
  long long n, x, y, c;
  while (scanf("%I64d%I64d%I64d%I64d", &n, &x, &y, &c) == 4) {
    long long es = 0, di = 2 * n;
    while (di > es) {
      long long me = (es + di) / 2;
      long long res = (me + 1) * (me + 1) + me * me;
      if (me - (n - x) > 0) res -= (me - (n - x)) * (me - (n - x));
      if (me - x + 1 > 0) res -= (me - x + 1) * (me - x + 1);
      if (me - (n - y) > 0) res -= (me - (n - y)) * (me - (n - y));
      if (me - y + 1 > 0) res -= (me - y + 1) * (me - y + 1);
      if (me - (n - x) - y > 0) res += foo(me - (n - x) - y);
      if (me - (n - x) - (n - y) - 1 > 0)
        res += foo(me - (n - x) - (n - y) - 1);
      if (me - x + 1 - y > 0) res += foo(me - x + 1 - y);
      if (me - x - (n - y) > 0) res += foo(me - x - (n - y));
      if (res >= c)
        di = me;
      else
        es = me + 1;
    }
    printf("%I64d\n", es);
  }
  return 0;
}

```