#include <bits/stdc++.h>
long long n, x, y, c;
long long calc(long long x, long long y, long long mid) {
  if (mid >= x + y - 2) return x * y;
  if (x > y) std::swap(x, y);
  if (mid < x) return (1 + mid + 1) * (mid + 1) / 2;
  if (mid < y) {
    long long ret = (1 + x) * x / 2;
    ret += (mid + 1 - x) * x;
    return ret;
  }
  long long t = x - 1 - (mid + 1 - y);
  return x * y - (1 + t) * t / 2;
}
bool judge(long long mid) {
  long long count = 1;
  count += calc(x, y, mid);
  count += calc(x, n - y + 1, mid);
  count += calc(n - x + 1, y, mid);
  count += calc(n - x + 1, n - y + 1, mid);
  count -= std::min(mid + 1, x);
  count -= std::min(mid + 1, y);
  count -= std::min(mid + 1, n - x + 1);
  count -= std::min(mid + 1, n - y + 1);
  return count >= c;
}
int main() {
  std::ios_base::sync_with_stdio(false);
  std::cin >> n >> x >> y >> c;
  long long l = 0, r = 4 * n;
  while (l + 1 < r) {
    long long mid = (l + r) / 2;
    if (judge(mid))
      r = mid;
    else
      l = mid;
  }
  if (judge(l)) r = l;
  std::cout << r << "\n";
}

### 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>
long long n, x, y, c;
long long calc(long long x, long long y, long long mid) {
  if (mid >= x + y - 2) return x * y;
  if (x > y) std::swap(x, y);
  if (mid < x) return (1 + mid + 1) * (mid + 1) / 2;
  if (mid < y) {
    long long ret = (1 + x) * x / 2;
    ret += (mid + 1 - x) * x;
    return ret;
  }
  long long t = x - 1 - (mid + 1 - y);
  return x * y - (1 + t) * t / 2;
}
bool judge(long long mid) {
  long long count = 1;
  count += calc(x, y, mid);
  count += calc(x, n - y + 1, mid);
  count += calc(n - x + 1, y, mid);
  count += calc(n - x + 1, n - y + 1, mid);
  count -= std::min(mid + 1, x);
  count -= std::min(mid + 1, y);
  count -= std::min(mid + 1, n - x + 1);
  count -= std::min(mid + 1, n - y + 1);
  return count >= c;
}
int main() {
  std::ios_base::sync_with_stdio(false);
  std::cin >> n >> x >> y >> c;
  long long l = 0, r = 4 * n;
  while (l + 1 < r) {
    long long mid = (l + r) / 2;
    if (judge(mid))
      r = mid;
    else
      l = mid;
  }
  if (judge(l)) r = l;
  std::cout << r << "\n";
}

```