#include <bits/stdc++.h>
using namespace std;
long long n, x, y, c, l, r, mid;
long long cal1(long long n, long long x, long long y, long long mid) {
  long long xx = x - mid, ret = 0;
  if (xx >= 1) return 0;
  ret += (2 - 2 * xx) * (-xx + 1) / 2;
  return ret;
}
long long cal2(long long n, long long x, long long y, long long mid) {
  mid -= x + y;
  if (mid < 0) return 0;
  return (mid + 1) * (mid + 2) / 2;
}
long long count(long long n, long long x, long long y, long long mid) {
  long long ret = 0;
  ret = 2 * mid * (mid + 1) + 1;
  for (int i = int(0); i < int(4); ++i) {
    ret -= cal1(n, x, y, mid);
    ret += cal2(n, x, y, mid);
    swap(x, y);
    y = n - y + 1;
  }
  return ret;
}
int main() {
  cin >> n >> x >> y >> c;
  l = 0;
  r = 2 * n;
  count(n, x, y, 4);
  while (l != r) {
    mid = (l + r) / 2;
    if (count(n, x, y, mid) >= c)
      r = mid;
    else
      l = mid + 1;
  }
  cout << l << 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, l, r, mid;
long long cal1(long long n, long long x, long long y, long long mid) {
  long long xx = x - mid, ret = 0;
  if (xx >= 1) return 0;
  ret += (2 - 2 * xx) * (-xx + 1) / 2;
  return ret;
}
long long cal2(long long n, long long x, long long y, long long mid) {
  mid -= x + y;
  if (mid < 0) return 0;
  return (mid + 1) * (mid + 2) / 2;
}
long long count(long long n, long long x, long long y, long long mid) {
  long long ret = 0;
  ret = 2 * mid * (mid + 1) + 1;
  for (int i = int(0); i < int(4); ++i) {
    ret -= cal1(n, x, y, mid);
    ret += cal2(n, x, y, mid);
    swap(x, y);
    y = n - y + 1;
  }
  return ret;
}
int main() {
  cin >> n >> x >> y >> c;
  l = 0;
  r = 2 * n;
  count(n, x, y, 4);
  while (l != r) {
    mid = (l + r) / 2;
    if (count(n, x, y, mid) >= c)
      r = mid;
    else
      l = mid + 1;
  }
  cout << l << endl;
  return 0;
}

```