#include <bits/stdc++.h>
using namespace std;
long long n, x, y, c;
long long c2(long long n) { return n * (n + 1) / 2; }
long long cnt(long long t) {
  long long sum = c2(t + 1) * 4 - 4 * (t + 1) + 1;
  if (t >= x) sum -= c2(t - x + 1) * 2 - (t - x + 1);
  if (t > n - x) sum -= c2(t - n + x) * 2 - (t - n + x);
  if (t >= y) sum -= c2(t - y + 1) * 2 - (t - y + 1);
  if (t > n - y) sum -= c2(t - n + y) * 2 - (t - n + y);
  if (t >= x + y) sum += c2(t - x - y + 1);
  if (t >= n - x + 1 + y) sum += c2(t - n + x - y);
  if (t >= n - y + 1 + x) sum += c2(t - n + y - x);
  if (t >= n - y + n - x + 2) sum += c2(t - n + y - n + x - 1);
  return sum;
}
long long ans(long long min, long long max) {
  if (min == max) return min;
  long long mid = (min + max) / 2;
  if (cnt(mid) >= c)
    return ans(min, mid);
  else
    return ans(mid + 1, max);
}
int main() {
  cin >> n >> x >> y >> c;
  cout << ans(0, 2 * n);
  return 0;
}

### 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;
long long c2(long long n) { return n * (n + 1) / 2; }
long long cnt(long long t) {
  long long sum = c2(t + 1) * 4 - 4 * (t + 1) + 1;
  if (t >= x) sum -= c2(t - x + 1) * 2 - (t - x + 1);
  if (t > n - x) sum -= c2(t - n + x) * 2 - (t - n + x);
  if (t >= y) sum -= c2(t - y + 1) * 2 - (t - y + 1);
  if (t > n - y) sum -= c2(t - n + y) * 2 - (t - n + y);
  if (t >= x + y) sum += c2(t - x - y + 1);
  if (t >= n - x + 1 + y) sum += c2(t - n + x - y);
  if (t >= n - y + 1 + x) sum += c2(t - n + y - x);
  if (t >= n - y + n - x + 2) sum += c2(t - n + y - n + x - 1);
  return sum;
}
long long ans(long long min, long long max) {
  if (min == max) return min;
  long long mid = (min + max) / 2;
  if (cnt(mid) >= c)
    return ans(min, mid);
  else
    return ans(mid + 1, max);
}
int main() {
  cin >> n >> x >> y >> c;
  cout << ans(0, 2 * n);
  return 0;
}

```