#include <bits/stdc++.h>
using namespace std;
const long long MOD = 1e9 + 7;
long long n, x, y, c, lb, ub, mid;
inline long long sum(long long a) { return a * (a + 1) / 2; }
inline bool check() {
  long long ret = ((mid) * (mid)) + ((mid + 1) * (mid + 1)) -
                  ((max(0LL, mid - x + 1)) * (max(0LL, mid - x + 1))) -
                  ((max(0LL, mid - n + x)) * (max(0LL, mid - n + x))) -
                  ((max(0LL, mid - y + 1)) * (max(0LL, mid - y + 1))) -
                  ((max(0LL, mid - n + y)) * (max(0LL, mid - n + y))) +
                  sum(max(0LL, mid - x - (n - y + 1) + 1)) +
                  sum(max(0LL, mid - x - y + 1)) +
                  sum(max(0LL, mid - (n - x + 1) - y + 1)) +
                  sum(max(0LL, mid - (n - x + 1) - (n - y + 1) + 1));
  return ret >= c;
}
int main() {
  cin >> n >> x >> y >> c;
  lb = 0;
  ub = 1e8;
  while (lb < ub) {
    mid = (lb + ub) / 2;
    if (check())
      ub = mid;
    else
      lb = mid + 1;
  }
  cout << lb << endl;
  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;
const long long MOD = 1e9 + 7;
long long n, x, y, c, lb, ub, mid;
inline long long sum(long long a) { return a * (a + 1) / 2; }
inline bool check() {
  long long ret = ((mid) * (mid)) + ((mid + 1) * (mid + 1)) -
                  ((max(0LL, mid - x + 1)) * (max(0LL, mid - x + 1))) -
                  ((max(0LL, mid - n + x)) * (max(0LL, mid - n + x))) -
                  ((max(0LL, mid - y + 1)) * (max(0LL, mid - y + 1))) -
                  ((max(0LL, mid - n + y)) * (max(0LL, mid - n + y))) +
                  sum(max(0LL, mid - x - (n - y + 1) + 1)) +
                  sum(max(0LL, mid - x - y + 1)) +
                  sum(max(0LL, mid - (n - x + 1) - y + 1)) +
                  sum(max(0LL, mid - (n - x + 1) - (n - y + 1) + 1));
  return ret >= c;
}
int main() {
  cin >> n >> x >> y >> c;
  lb = 0;
  ub = 1e8;
  while (lb < ub) {
    mid = (lb + ub) / 2;
    if (check())
      ub = mid;
    else
      lb = mid + 1;
  }
  cout << lb << endl;
  return 0;
}

```