#include <bits/stdc++.h>
using namespace std;
long long n, x, y, c, l = -1, r = 1e12;
long long calc(long long n, long long m, long long k) {
  if (n > m) swap(n, m);
  if (k <= 0 || n <= 0) return 0;
  if (k <= n) return (k + 1) * k / 2;
  return min(k - n, m) * n + calc(n, m + n - k, n);
}
long long check(long long t) {
  return calc(x, y, t + 1) + calc(n - x, n - y, t - 1) + calc(n - x, y, t) +
         calc(x, n - y, t);
}
int main() {
  cin >> n >> x >> y >> c;
  while (l < r - 1) (check(l + r >> 1) >= c ? r : l) = l + r >> 1;
  cout << r;
}

### Prompt

Create a solution in CPP for 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 n, x, y, c, l = -1, r = 1e12;
long long calc(long long n, long long m, long long k) {
  if (n > m) swap(n, m);
  if (k <= 0 || n <= 0) return 0;
  if (k <= n) return (k + 1) * k / 2;
  return min(k - n, m) * n + calc(n, m + n - k, n);
}
long long check(long long t) {
  return calc(x, y, t + 1) + calc(n - x, n - y, t - 1) + calc(n - x, y, t) +
         calc(x, n - y, t);
}
int main() {
  cin >> n >> x >> y >> c;
  while (l < r - 1) (check(l + r >> 1) >= c ? r : l) = l + r >> 1;
  cout << r;
}

```