#include <bits/stdc++.h>
using namespace std;
long long n, x, y, c;
long long f(long long p, long long q, long long r) {
  r--;
  if (p > q) swap(p, q);
  if (r >= p + q - 1) return p * q;
  long long cnt1 = r * (r + 1) / 2;
  if (r <= p) return cnt1;
  long long cnt2 = p * (p + 1) / 2 + (r - p) * p;
  if (r <= q) return cnt2;
  long long t = p + q - 1 - r;
  return p * q - t * (t + 1) / 2;
}
bool judge(long long z) {
  long long tot = f(x - 1, y - 1, z) + f(n - x, y - 1, z) + f(x - 1, n - y, z) +
                  f(n - x, n - y, z);
  tot += min(x - 1, z) + min(n - x, z) + min(y - 1, z) + min(n - y, z) + 1;
  return tot >= c;
}
int main() {
  cin >> n >> x >> y >> c;
  long long l = 0, r = 2 * n, ans = 0;
  while (l <= r) {
    long long mid = (l + r) >> 1;
    if (judge(mid)) {
      ans = mid;
      r = mid - 1;
    } else
      l = mid + 1;
  }
  cout << ans << endl;
  return 0;
}

### Prompt

Generate 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;
long long n, x, y, c;
long long f(long long p, long long q, long long r) {
  r--;
  if (p > q) swap(p, q);
  if (r >= p + q - 1) return p * q;
  long long cnt1 = r * (r + 1) / 2;
  if (r <= p) return cnt1;
  long long cnt2 = p * (p + 1) / 2 + (r - p) * p;
  if (r <= q) return cnt2;
  long long t = p + q - 1 - r;
  return p * q - t * (t + 1) / 2;
}
bool judge(long long z) {
  long long tot = f(x - 1, y - 1, z) + f(n - x, y - 1, z) + f(x - 1, n - y, z) +
                  f(n - x, n - y, z);
  tot += min(x - 1, z) + min(n - x, z) + min(y - 1, z) + min(n - y, z) + 1;
  return tot >= c;
}
int main() {
  cin >> n >> x >> y >> c;
  long long l = 0, r = 2 * n, ans = 0;
  while (l <= r) {
    long long mid = (l + r) >> 1;
    if (judge(mid)) {
      ans = mid;
      r = mid - 1;
    } else
      l = mid + 1;
  }
  cout << ans << endl;
  return 0;
}

```