#include <bits/stdc++.h>
using namespace std;
template <typename T>
T sqr(T x) {
  return x * x;
}
template <typename T>
T Abs(T x) {
  return x > 0 ? x : -x;
}
long long c, n, x, y, l, r, mid, ans;
int main() {
  cin >> n >> x >> y >> c;
  l = 0;
  r = n + n - 2;
  while (l <= r) {
    mid = (l + r) / 2;
    long long tot = 0;
    if (x - mid <= 0) tot += sqr(Abs(x - mid - 1));
    if (x - mid <= 0 && y - mid <= 0) {
      long long res = mid - (x - 1) - (y - 1) - 1;
      if (res > 0) tot -= (sqr(res) + res) / 2;
    }
    if (x + mid > n) tot += sqr(x + mid - n);
    if (x - mid <= 0 && y + mid > n) {
      long long res = mid - (x - 1) - (n - y) - 1;
      if (res > 0) tot -= (sqr(res) + res) / 2;
    }
    if (y - mid <= 0) tot += sqr(Abs(y - mid - 1));
    if (x + mid > n && y - mid <= 0) {
      long long res = mid - (n - x) - (y - 1) - 1;
      if (res > 0) tot -= (sqr(res) + res) / 2;
    }
    if (y + mid > n) tot += sqr(y + mid - n);
    if (x + mid > n && y + mid > n) {
      long long res = mid - (n - x) - (n - y) - 1;
      if (res > 0) tot -= (sqr(res) + res) / 2;
    }
    tot = sqr(mid) + sqr(mid + 1) - tot;
    if (tot >= c)
      ans = mid, r = mid - 1;
    else
      l = mid + 1;
  }
  cout << ans << endl;
  return 0;
}

### 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;
template <typename T>
T sqr(T x) {
  return x * x;
}
template <typename T>
T Abs(T x) {
  return x > 0 ? x : -x;
}
long long c, n, x, y, l, r, mid, ans;
int main() {
  cin >> n >> x >> y >> c;
  l = 0;
  r = n + n - 2;
  while (l <= r) {
    mid = (l + r) / 2;
    long long tot = 0;
    if (x - mid <= 0) tot += sqr(Abs(x - mid - 1));
    if (x - mid <= 0 && y - mid <= 0) {
      long long res = mid - (x - 1) - (y - 1) - 1;
      if (res > 0) tot -= (sqr(res) + res) / 2;
    }
    if (x + mid > n) tot += sqr(x + mid - n);
    if (x - mid <= 0 && y + mid > n) {
      long long res = mid - (x - 1) - (n - y) - 1;
      if (res > 0) tot -= (sqr(res) + res) / 2;
    }
    if (y - mid <= 0) tot += sqr(Abs(y - mid - 1));
    if (x + mid > n && y - mid <= 0) {
      long long res = mid - (n - x) - (y - 1) - 1;
      if (res > 0) tot -= (sqr(res) + res) / 2;
    }
    if (y + mid > n) tot += sqr(y + mid - n);
    if (x + mid > n && y + mid > n) {
      long long res = mid - (n - x) - (n - y) - 1;
      if (res > 0) tot -= (sqr(res) + res) / 2;
    }
    tot = sqr(mid) + sqr(mid + 1) - tot;
    if (tot >= c)
      ans = mid, r = mid - 1;
    else
      l = mid + 1;
  }
  cout << ans << endl;
  return 0;
}

```