#include <bits/stdc++.h>
using namespace std;
int main() {
  long long n, x, y, c;
  while (cin >> n >> x >> y >> c) {
    long long num = 1;
    long long ans = -1;
    for (long long step = 1;; step++) {
      if (num >= c) {
        ans = step - 1;
        break;
      }
      long long dup = min(step, abs(x - 1));
      long long ddown = min(step, abs(x - n));
      long long dleft = min(step, abs(y - 1));
      long long dright = min(step, abs(y - n));
      if (dup == step) num++;
      if (ddown == step) num++;
      if (dleft == step) num++;
      if (dright == step) num++;
      long long t1, t2;
      t1 = min(dup, step - 1);
      t2 = min(dleft, step - 1);
      num += max(0LL, t1 + t2 - step + 1);
      t1 = min(dup, step - 1);
      t2 = min(dright, step - 1);
      num += max(0LL, t1 + t2 - step + 1);
      t1 = min(ddown, step - 1);
      t2 = min(dleft, step - 1);
      num += max(0LL, t1 + t2 - step + 1);
      t1 = min(ddown, step - 1);
      t2 = min(dright, step - 1);
      num += max(0LL, t1 + t2 - step + 1);
    }
    cout << ans << endl;
  }
  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;
int main() {
  long long n, x, y, c;
  while (cin >> n >> x >> y >> c) {
    long long num = 1;
    long long ans = -1;
    for (long long step = 1;; step++) {
      if (num >= c) {
        ans = step - 1;
        break;
      }
      long long dup = min(step, abs(x - 1));
      long long ddown = min(step, abs(x - n));
      long long dleft = min(step, abs(y - 1));
      long long dright = min(step, abs(y - n));
      if (dup == step) num++;
      if (ddown == step) num++;
      if (dleft == step) num++;
      if (dright == step) num++;
      long long t1, t2;
      t1 = min(dup, step - 1);
      t2 = min(dleft, step - 1);
      num += max(0LL, t1 + t2 - step + 1);
      t1 = min(dup, step - 1);
      t2 = min(dright, step - 1);
      num += max(0LL, t1 + t2 - step + 1);
      t1 = min(ddown, step - 1);
      t2 = min(dleft, step - 1);
      num += max(0LL, t1 + t2 - step + 1);
      t1 = min(ddown, step - 1);
      t2 = min(dright, step - 1);
      num += max(0LL, t1 + t2 - step + 1);
    }
    cout << ans << endl;
  }
  return 0;
}

```