#include <bits/stdc++.h>
using namespace std;
long long ai00(long long m, long long x, long long y) {
  long long ans = 0;
  if (m > x) {
    long long t = min(m - x, y);
    ans = t * x;
    m -= t;
    y -= t;
  }
  y = min(m, y);
  ans += (m + m - y + 1) * y / 2;
  return ans;
}
long long ai00(long long m, long long n, long long x, long long y) {
  long long ans = 1;
  ans += ai00(m, x, n - y);
  ans += ai00(m, y, x - 1);
  ans += ai00(m, n - x + 1, y - 1);
  ans += ai00(m, n - y + 1, n - x);
  return ans;
}
int main() {
  long long n, x, y;
  long long c;
  while (cin >> n >> x >> y >> c) {
    int ans;
    long long s = 0;
    long long t = 3 * n;
    while (s <= t) {
      long long mid = (s + t) >> 1;
      if (ai00(mid, n, x, y) >= c) {
        ans = mid;
        t = mid - 1;
      } else
        s = mid + 1;
    }
    cout << ans << endl;
  }
}

### Prompt

Construct a Cpp code solution to the problem outlined:

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 ai00(long long m, long long x, long long y) {
  long long ans = 0;
  if (m > x) {
    long long t = min(m - x, y);
    ans = t * x;
    m -= t;
    y -= t;
  }
  y = min(m, y);
  ans += (m + m - y + 1) * y / 2;
  return ans;
}
long long ai00(long long m, long long n, long long x, long long y) {
  long long ans = 1;
  ans += ai00(m, x, n - y);
  ans += ai00(m, y, x - 1);
  ans += ai00(m, n - x + 1, y - 1);
  ans += ai00(m, n - y + 1, n - x);
  return ans;
}
int main() {
  long long n, x, y;
  long long c;
  while (cin >> n >> x >> y >> c) {
    int ans;
    long long s = 0;
    long long t = 3 * n;
    while (s <= t) {
      long long mid = (s + t) >> 1;
      if (ai00(mid, n, x, y) >= c) {
        ans = mid;
        t = mid - 1;
      } else
        s = mid + 1;
    }
    cout << ans << endl;
  }
}

```