#include <bits/stdc++.h>
using namespace std;
long long n, x, y, c;
long long mx[4];
long long my[4];
long long NUM(long long x, long long y, long long t) {
  if (x * y == 0) return x * y;
  if (t >= x + y - 1) return x * y;
  if (x > y) swap(x, y);
  if (t <= x) return t * (t + 1) / 2;
  if (t <= x + y - x - 1) return x * (x + 1) / 2 + (t - x) * x;
  long long tmp = x + y - 1 - t;
  return x * y - tmp * (tmp + 1) / 2;
}
long long cal(long long t) {
  long long ans = 1;
  if (t == 0) return 1;
  for (int i = 0; i < 4; i++) {
    ans += NUM(mx[i], my[i], t - 1);
  }
  ans += min(x - 1, t);
  ans += min(n - x, t);
  ans += min(y - 1, t);
  ans += min(n - y, t);
  return ans;
}
void init() {
  cin >> n >> x >> y >> c;
  mx[0] = x - 1;
  my[0] = y - 1;
  mx[1] = x - 1;
  my[1] = n - y;
  mx[2] = n - x;
  my[2] = y - 1;
  mx[3] = n - x;
  my[3] = n - y;
  long long s = 0, t = 1e9;
  while (s <= t) {
    long long mid = (s + t) >> 1;
    if (cal(mid) < c)
      s = mid + 1;
    else
      t = mid - 1;
  }
  cout << s << endl;
}
int main() {
  init();
  return 0;
}

### Prompt

Construct a cpp code solution to the problem outlined:

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 mx[4];
long long my[4];
long long NUM(long long x, long long y, long long t) {
  if (x * y == 0) return x * y;
  if (t >= x + y - 1) return x * y;
  if (x > y) swap(x, y);
  if (t <= x) return t * (t + 1) / 2;
  if (t <= x + y - x - 1) return x * (x + 1) / 2 + (t - x) * x;
  long long tmp = x + y - 1 - t;
  return x * y - tmp * (tmp + 1) / 2;
}
long long cal(long long t) {
  long long ans = 1;
  if (t == 0) return 1;
  for (int i = 0; i < 4; i++) {
    ans += NUM(mx[i], my[i], t - 1);
  }
  ans += min(x - 1, t);
  ans += min(n - x, t);
  ans += min(y - 1, t);
  ans += min(n - y, t);
  return ans;
}
void init() {
  cin >> n >> x >> y >> c;
  mx[0] = x - 1;
  my[0] = y - 1;
  mx[1] = x - 1;
  my[1] = n - y;
  mx[2] = n - x;
  my[2] = y - 1;
  mx[3] = n - x;
  my[3] = n - y;
  long long s = 0, t = 1e9;
  while (s <= t) {
    long long mid = (s + t) >> 1;
    if (cal(mid) < c)
      s = mid + 1;
    else
      t = mid - 1;
  }
  cout << s << endl;
}
int main() {
  init();
  return 0;
}

```