#include <bits/stdc++.h>
using namespace std;
long long n, x, y, c;
long long blocks(long long t) {
  long long res = 0;
  long long a1 = -1, a2 = 1, f = y, s = y;
  long long midu = x - t, midd = x + t, midl = y - t, midr = y + t;
  for (long long i = midu; i <= midd; i++) {
    if (i > 0 && i <= n) res += min(s, n) - max(1ll, f) + 1;
    if (f == midl) swap(a1, a2);
    f += a1, s += a2;
  }
  return res;
}
int main() {
  cin >> n >> x >> y >> c;
  long long l = 0, r = 1e5;
  while (r > l) {
    long long mid = (l + r) / 2;
    if (blocks(mid) >= c)
      r = mid;
    else
      l = mid + 1;
  }
  cout << r;
  return 0;
}

### Prompt

In cpp, your task is to solve 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;
long long blocks(long long t) {
  long long res = 0;
  long long a1 = -1, a2 = 1, f = y, s = y;
  long long midu = x - t, midd = x + t, midl = y - t, midr = y + t;
  for (long long i = midu; i <= midd; i++) {
    if (i > 0 && i <= n) res += min(s, n) - max(1ll, f) + 1;
    if (f == midl) swap(a1, a2);
    f += a1, s += a2;
  }
  return res;
}
int main() {
  cin >> n >> x >> y >> c;
  long long l = 0, r = 1e5;
  while (r > l) {
    long long mid = (l + r) / 2;
    if (blocks(mid) >= c)
      r = mid;
    else
      l = mid + 1;
  }
  cout << r;
  return 0;
}

```