#include <bits/stdc++.h>
using namespace std;
long long a, b, c, d;
long long sum;
int main() {
  while (scanf("%I64d%I64d%I64d%I64d", &a, &b, &c, &d) == 4) {
    long long up = b - 1;
    long long down = a - b;
    long long left = c - 1;
    long long right = a - c;
    for (long long i = 0;; ++i) {
      sum = 2 * i * i + 2 * i + 1;
      if (i > up) sum -= (i - up) * (i - up);
      if (i > down) sum -= (i - down) * (i - down);
      if (i > left) sum -= (i - left) * (i - left);
      if (i > right) sum -= (i - right) * (i - right);
      if (i > up + right)
        sum += (i - (up + right + 1)) * (i - (up + right + 1) + 1) / 2;
      if (i > up + left)
        sum += (i - (up + left + 1)) * (i - (up + left + 1) + 1) / 2;
      if (i > down + left)
        sum += (i - (down + left + 1)) * (i - (down + left + 1) + 1) / 2;
      if (i > down + right)
        sum += (i - (down + right + 1)) * (i - (down + right + 1) + 1) / 2;
      if (sum >= d) {
        cout << i << endl;
        break;
      }
    }
  }
  return 0;
}

### Prompt

Please create a solution in CPP 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 a, b, c, d;
long long sum;
int main() {
  while (scanf("%I64d%I64d%I64d%I64d", &a, &b, &c, &d) == 4) {
    long long up = b - 1;
    long long down = a - b;
    long long left = c - 1;
    long long right = a - c;
    for (long long i = 0;; ++i) {
      sum = 2 * i * i + 2 * i + 1;
      if (i > up) sum -= (i - up) * (i - up);
      if (i > down) sum -= (i - down) * (i - down);
      if (i > left) sum -= (i - left) * (i - left);
      if (i > right) sum -= (i - right) * (i - right);
      if (i > up + right)
        sum += (i - (up + right + 1)) * (i - (up + right + 1) + 1) / 2;
      if (i > up + left)
        sum += (i - (up + left + 1)) * (i - (up + left + 1) + 1) / 2;
      if (i > down + left)
        sum += (i - (down + left + 1)) * (i - (down + left + 1) + 1) / 2;
      if (i > down + right)
        sum += (i - (down + right + 1)) * (i - (down + right + 1) + 1) / 2;
      if (sum >= d) {
        cout << i << endl;
        break;
      }
    }
  }
  return 0;
}

```