#include <bits/stdc++.h>
using namespace std;
int main() {
  int i, j, n, x, y, m;
  long long k;
  scanf("%d%d%d%lld", &n, &x, &y, &k);
  int l = 0, r = 2 * n, mid;
  long long ans;
  while (l <= r) {
    mid = (l + r) >> 1;
    ans = (long long)mid * mid * 2 + 2 * mid + 1;
    if (x + mid - n >= 0) {
      ans -= (long long)(x + mid - n) * (x + mid - n);
    }
    if (x - mid <= 0) {
      ans -= (long long)(mid - x + 1) * (mid - x + 1);
    }
    if (y + mid - n >= 0) {
      ans -= (long long)(y + mid - n) * (y + mid - n);
    }
    if (y - mid <= 0) {
      ans -= (long long)(mid - y + 1) * (mid - y + 1);
    }
    if (mid >= x + y) {
      ans += (long long)(mid - x - y + 1) * (mid - x - y + 2) / 2;
    }
    if (mid >= n - y + x + 1) {
      ans += (long long)(mid - x - n + y) * (mid - x - n + y + 1) / 2;
    }
    if (mid >= n + y - x + 1) {
      ans += (long long)(mid - y - n + x) * (mid - y - n + x + 1) / 2;
    }
    if (mid >= n - x + n - y + 2) {
      ans += (long long)(mid - n + x - n + y - 1) * (mid - n + x - n + y) / 2;
    }
    if (ans >= k) {
      r = mid - 1;
    } else {
      l = mid + 1;
    }
  }
  printf("%d", r + 1);
  return 0;
}

### 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;
int main() {
  int i, j, n, x, y, m;
  long long k;
  scanf("%d%d%d%lld", &n, &x, &y, &k);
  int l = 0, r = 2 * n, mid;
  long long ans;
  while (l <= r) {
    mid = (l + r) >> 1;
    ans = (long long)mid * mid * 2 + 2 * mid + 1;
    if (x + mid - n >= 0) {
      ans -= (long long)(x + mid - n) * (x + mid - n);
    }
    if (x - mid <= 0) {
      ans -= (long long)(mid - x + 1) * (mid - x + 1);
    }
    if (y + mid - n >= 0) {
      ans -= (long long)(y + mid - n) * (y + mid - n);
    }
    if (y - mid <= 0) {
      ans -= (long long)(mid - y + 1) * (mid - y + 1);
    }
    if (mid >= x + y) {
      ans += (long long)(mid - x - y + 1) * (mid - x - y + 2) / 2;
    }
    if (mid >= n - y + x + 1) {
      ans += (long long)(mid - x - n + y) * (mid - x - n + y + 1) / 2;
    }
    if (mid >= n + y - x + 1) {
      ans += (long long)(mid - y - n + x) * (mid - y - n + x + 1) / 2;
    }
    if (mid >= n - x + n - y + 2) {
      ans += (long long)(mid - n + x - n + y - 1) * (mid - n + x - n + y) / 2;
    }
    if (ans >= k) {
      r = mid - 1;
    } else {
      l = mid + 1;
    }
  }
  printf("%d", r + 1);
  return 0;
}

```