#include <bits/stdc++.h>
using namespace std;
long long n, r, c, x;
long long max(long long a, long long b) { return a > b ? a : b; }
long long total(long long m) {
  long long cnt = 0;
  for (long long i = max(1, r - m); i <= min(n, r + m); i++) {
    long long L, R;
    L = c - m + abs(r - i);
    R = c + m - abs(r - i);
    cnt += max(0, 1 + min(n, R) - max(1, L));
  }
  return cnt;
}
long long fn() {
  long long l = 0, h = 1e7, m;
  for (long long i = 0; i < 200; i++) {
    m = (l + h) / 2;
    long long p = total(m);
    if (h - l <= 1) break;
    if (p >= x)
      h = m;
    else
      l = m;
  }
  return total(l) >= x ? l : h;
}
int main() {
  int t, tc = 0;
  cin >> n >> r >> c >> x;
  long long ans = fn();
  cout << ans << 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;
long long n, r, c, x;
long long max(long long a, long long b) { return a > b ? a : b; }
long long total(long long m) {
  long long cnt = 0;
  for (long long i = max(1, r - m); i <= min(n, r + m); i++) {
    long long L, R;
    L = c - m + abs(r - i);
    R = c + m - abs(r - i);
    cnt += max(0, 1 + min(n, R) - max(1, L));
  }
  return cnt;
}
long long fn() {
  long long l = 0, h = 1e7, m;
  for (long long i = 0; i < 200; i++) {
    m = (l + h) / 2;
    long long p = total(m);
    if (h - l <= 1) break;
    if (p >= x)
      h = m;
    else
      l = m;
  }
  return total(l) >= x ? l : h;
}
int main() {
  int t, tc = 0;
  cin >> n >> r >> c >> x;
  long long ans = fn();
  cout << ans << endl;
  return 0;
}

```