#include <bits/stdc++.h>
using namespace std;
int n, x, y;
long long cal1(int lim, long long len) {
  if (len <= lim) return 0;
  len -= lim;
  return len * len;
}
long long cal2(int lim1, int lim2, long long len) {
  len -= lim1 + lim2;
  if (len <= 0) return 0;
  return (1 + len) * len / 2;
}
long long check(long long len) {
  long long ret = len * len + (len + 1) * (len + 1);
  ret -= cal1(x, len + 1);
  ret -= cal1(n - x + 1, len + 1);
  ret -= cal1(n - y + 1, len + 1);
  ret -= cal1(y, len + 1);
  ret += cal2(x, n - y + 1, len + 1);
  ret += cal2(n - y + 1, n - x + 1, len + 1);
  ret += cal2(n - x + 1, y, len + 1);
  ret += cal2(x, y, len + 1);
  return ret;
}
int main() {
  while (scanf("%d", &n) == 1) {
    int l = 0, r = 2 * n, mid;
    long long c;
    scanf("%d%d", &x, &y);
    cin >> c;
    while (l < r) {
      mid = l + (r - l) / 2;
      if (check(mid) >= c)
        r = mid;
      else
        l = mid + 1;
    }
    printf("%d\n", l);
  }
  return 0;
}

### Prompt

Please provide a CPP coded solution to the problem described below:

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 cal1(int lim, long long len) {
  if (len <= lim) return 0;
  len -= lim;
  return len * len;
}
long long cal2(int lim1, int lim2, long long len) {
  len -= lim1 + lim2;
  if (len <= 0) return 0;
  return (1 + len) * len / 2;
}
long long check(long long len) {
  long long ret = len * len + (len + 1) * (len + 1);
  ret -= cal1(x, len + 1);
  ret -= cal1(n - x + 1, len + 1);
  ret -= cal1(n - y + 1, len + 1);
  ret -= cal1(y, len + 1);
  ret += cal2(x, n - y + 1, len + 1);
  ret += cal2(n - y + 1, n - x + 1, len + 1);
  ret += cal2(n - x + 1, y, len + 1);
  ret += cal2(x, y, len + 1);
  return ret;
}
int main() {
  while (scanf("%d", &n) == 1) {
    int l = 0, r = 2 * n, mid;
    long long c;
    scanf("%d%d", &x, &y);
    cin >> c;
    while (l < r) {
      mid = l + (r - l) / 2;
      if (check(mid) >= c)
        r = mid;
      else
        l = mid + 1;
    }
    printf("%d\n", l);
  }
  return 0;
}

```