#include <bits/stdc++.h>
using namespace std;
const int maxn = 1000100;
const int INF = 1e9 + 10;
long long n, x, y, c;
long long g(long long k) {
  if (k <= 0) return 0;
  return k * k;
}
long long h(long long k) {
  if (k <= 0) return 0;
  return k * (k + 1) / 2;
}
long long f(long long k) {
  if (k < 0) return 0;
  return 1 + 2 * k * (k + 1) - g(y + k - n) - g(x + k - n) - g(1 - (y - k)) -
         g(1 - (x - k)) + h(y + k - n - x) + h(x + k - n - y) +
         h(y + k - n - (n - x + 1)) + h(1 - (y - k) - x);
}
long long bin(long long l, long long r, long long x) {
  long long res = r;
  while (l <= r) {
    long long m = (l + r) >> 1;
    long long F = f(m);
    if (F >= x)
      res = min(res, m), r = m - 1;
    else
      l = m + 1;
  }
  return res;
}
int main() {
  while (cin >> n >> x >> y >> c) {
    cout << bin(0, n * 3, c) << 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 int maxn = 1000100;
const int INF = 1e9 + 10;
long long n, x, y, c;
long long g(long long k) {
  if (k <= 0) return 0;
  return k * k;
}
long long h(long long k) {
  if (k <= 0) return 0;
  return k * (k + 1) / 2;
}
long long f(long long k) {
  if (k < 0) return 0;
  return 1 + 2 * k * (k + 1) - g(y + k - n) - g(x + k - n) - g(1 - (y - k)) -
         g(1 - (x - k)) + h(y + k - n - x) + h(x + k - n - y) +
         h(y + k - n - (n - x + 1)) + h(1 - (y - k) - x);
}
long long bin(long long l, long long r, long long x) {
  long long res = r;
  while (l <= r) {
    long long m = (l + r) >> 1;
    long long F = f(m);
    if (F >= x)
      res = min(res, m), r = m - 1;
    else
      l = m + 1;
  }
  return res;
}
int main() {
  while (cin >> n >> x >> y >> c) {
    cout << bin(0, n * 3, c) << endl;
  }
  return 0;
}

```