#include <bits/stdc++.h>
using namespace std;
long long n, x, y, c, lth;
long long cal(long long cnt, long long len) {
  if (cnt < 0) cnt = 0;
  long long ans = 0;
  long long nowl = lth;
  if (lth > len) {
    long long c = min(cnt, lth - len);
    ans += 1LL * len * c;
    cnt -= c;
    nowl = len;
  }
  ans = ans + 1LL * cnt * (2 * nowl - cnt + 1) / 2;
  return ans;
}
long long work(long long cnt) {
  long long ans =
      1 + min(cnt, y - 1) + min(cnt, n - y) + min(cnt, x - 1) + min(cnt, n - x);
  long long c1 = min(cnt - 1, x - 1);
  long long c2 = min(cnt - 1, n - x);
  lth = cnt - 1;
  ans = ans + cal(c1, y - 1) + cal(c2, y - 1) + cal(c1, n - y) + cal(c2, n - y);
  return ans;
}
int main() {
  cin >> n >> x >> y >> c;
  int l = 0, r = c;
  while (l < r) {
    long long mid = l + r >> 1;
    if (work(mid) >= c)
      r = mid;
    else
      l = mid + 1;
  }
  cout << r << endl;
}

### Prompt

Your task is to create 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, lth;
long long cal(long long cnt, long long len) {
  if (cnt < 0) cnt = 0;
  long long ans = 0;
  long long nowl = lth;
  if (lth > len) {
    long long c = min(cnt, lth - len);
    ans += 1LL * len * c;
    cnt -= c;
    nowl = len;
  }
  ans = ans + 1LL * cnt * (2 * nowl - cnt + 1) / 2;
  return ans;
}
long long work(long long cnt) {
  long long ans =
      1 + min(cnt, y - 1) + min(cnt, n - y) + min(cnt, x - 1) + min(cnt, n - x);
  long long c1 = min(cnt - 1, x - 1);
  long long c2 = min(cnt - 1, n - x);
  lth = cnt - 1;
  ans = ans + cal(c1, y - 1) + cal(c2, y - 1) + cal(c1, n - y) + cal(c2, n - y);
  return ans;
}
int main() {
  cin >> n >> x >> y >> c;
  int l = 0, r = c;
  while (l < r) {
    long long mid = l + r >> 1;
    if (work(mid) >= c)
      r = mid;
    else
      l = mid + 1;
  }
  cout << r << endl;
}

```