#include <bits/stdc++.h>
using namespace std;
int n, x, y;
long long c;
long long sqr(long long v) { return v * v; }
long long cal2(long long v) { return v * (v + 1) / 2; }
long long cal(long long t) {
  long long res = 2 * t * t + 2 * t + 1;
  t++;
  res -= sqr(max(0ll, t - x));
  res -= sqr(max(0ll, t - (n - x + 1)));
  res -= sqr(max(0ll, t - y));
  res -= sqr(max(0ll, t - (n - y + 1)));
  long long len = t - (n - y + 1);
  if (len > 0) {
    res += cal2(max(0ll, len - x));
    res += cal2(max(0ll, len - (n - x + 1)));
  }
  len = t - y;
  if (len > 0) {
    res += cal2(max(0ll, len - x));
    res += cal2(max(0ll, len - (n - x + 1)));
  }
  return res;
}
int main() {
  cin >> n >> x >> y >> c;
  long long l = 0, r = 2000000000ll, mid, res = 1000000007;
  while (l <= r) {
    mid = (l + r) / 2;
    if (cal(mid) >= c) {
      res = mid;
      r = mid - 1;
    } else
      l = mid + 1;
  }
  cout << res << endl;
  return 0;
}

### 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>
using namespace std;
int n, x, y;
long long c;
long long sqr(long long v) { return v * v; }
long long cal2(long long v) { return v * (v + 1) / 2; }
long long cal(long long t) {
  long long res = 2 * t * t + 2 * t + 1;
  t++;
  res -= sqr(max(0ll, t - x));
  res -= sqr(max(0ll, t - (n - x + 1)));
  res -= sqr(max(0ll, t - y));
  res -= sqr(max(0ll, t - (n - y + 1)));
  long long len = t - (n - y + 1);
  if (len > 0) {
    res += cal2(max(0ll, len - x));
    res += cal2(max(0ll, len - (n - x + 1)));
  }
  len = t - y;
  if (len > 0) {
    res += cal2(max(0ll, len - x));
    res += cal2(max(0ll, len - (n - x + 1)));
  }
  return res;
}
int main() {
  cin >> n >> x >> y >> c;
  long long l = 0, r = 2000000000ll, mid, res = 1000000007;
  while (l <= r) {
    mid = (l + r) / 2;
    if (cal(mid) >= c) {
      res = mid;
      r = mid - 1;
    } else
      l = mid + 1;
  }
  cout << res << endl;
  return 0;
}

```