#include <bits/stdc++.h>
using namespace std;
long long x, y;
long long n, c;
long long tr(long long t) { return t * t; }
long long p(long long mid) {
  long long on = (mid + 1LL) * (mid + 1LL) + mid * mid;
  if (mid > x - 1) on -= tr(mid - x + 1);
  if (mid > y - 1) on -= tr(mid - y + 1);
  if (mid + y > n) on -= tr(mid + y - n);
  if (mid + x > n) on -= tr(mid + x - n);
  if (mid > x + y - 1) {
    long long k = mid - x - y + 1;
    on += (k * (k + 1)) / 2;
  }
  if (mid > y + n - x) {
    long long k = mid - y - n + x;
    on += (k * (k + 1)) / 2;
  }
  if (mid > x + n - y) {
    long long k = mid - x - n + y;
    on += (k * (k + 1)) / 2;
  }
  if (mid > n + n - x - y + 1) {
    long long k = mid - n - n + x + y - 1;
    on += (k * (k + 1)) / 2;
  }
  return on;
}
int main() {
  cin >> n >> x >> y >> c;
  if (c == 1)
    cout << 0 << endl;
  else {
    long long top = 2LL * n, bot = 0, mid = (top + bot) / 2;
    while (top - bot > 1) {
      long long on = p(mid);
      if (on >= c)
        top = mid;
      else
        bot = mid;
      mid = (top + bot) / 2;
    }
    cout << top << endl;
  }
}

### Prompt

Your challenge is to write 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 x, y;
long long n, c;
long long tr(long long t) { return t * t; }
long long p(long long mid) {
  long long on = (mid + 1LL) * (mid + 1LL) + mid * mid;
  if (mid > x - 1) on -= tr(mid - x + 1);
  if (mid > y - 1) on -= tr(mid - y + 1);
  if (mid + y > n) on -= tr(mid + y - n);
  if (mid + x > n) on -= tr(mid + x - n);
  if (mid > x + y - 1) {
    long long k = mid - x - y + 1;
    on += (k * (k + 1)) / 2;
  }
  if (mid > y + n - x) {
    long long k = mid - y - n + x;
    on += (k * (k + 1)) / 2;
  }
  if (mid > x + n - y) {
    long long k = mid - x - n + y;
    on += (k * (k + 1)) / 2;
  }
  if (mid > n + n - x - y + 1) {
    long long k = mid - n - n + x + y - 1;
    on += (k * (k + 1)) / 2;
  }
  return on;
}
int main() {
  cin >> n >> x >> y >> c;
  if (c == 1)
    cout << 0 << endl;
  else {
    long long top = 2LL * n, bot = 0, mid = (top + bot) / 2;
    while (top - bot > 1) {
      long long on = p(mid);
      if (on >= c)
        top = mid;
      else
        bot = mid;
      mid = (top + bot) / 2;
    }
    cout << top << endl;
  }
}

```