#include <bits/stdc++.h>
using namespace std;
long long n, x, y, c;
long long l = 0, r = 3e9, ans;
bool check(long long mid, long long need) {
  long long c = 2 * mid * mid + 2 * mid + 1;
  if (mid - x > 0) c -= (mid - x) * (mid - x);
  if (mid - y > 0) c -= (mid - y) * (mid - y);
  if (mid - (n - 1 - x) > 0) c -= (mid - (n - 1 - x)) * (mid - (n - 1 - x));
  if (mid - (n - 1 - y) > 0) c -= (mid - (n - 1 - y)) * (mid - (n - 1 - y));
  if (mid - x - (n - y - 1) - 1 > 0)
    c += (1 + mid - x - (n - y - 1) - 1) * (mid - x - (n - y - 1) - 1) / 2;
  if (mid - (n - 1 - x) - (n - 1 - y) > 0)
    c += (1 + mid - (n - 1 - x) - (n - 1 - y) - 1) *
         (mid - (n - 1 - x) - (n - 1 - y) - 1) / 2;
  if (mid - (n - 1 - x) - y - 1 > 0)
    c += (1 + mid - (n - 1 - x) - y - 1) * (mid - (n - 1 - x) - y - 1) / 2;
  if (mid - y - x - 1 > 0) c += (1 + mid - y - x - 1) * (mid - y - x - 1) / 2;
  return c >= need;
}
int main() {
  scanf("%lld%lld%lld%lld", &n, &x, &y, &c);
  x--;
  y--;
  while (l <= r) {
    long long mid = (l + r) >> 1;
    if (check(mid, c)) {
      ans = mid;
      r = mid - 1;
    } else {
      l = mid + 1;
    }
  }
  printf("%lld", ans);
  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;
long long n, x, y, c;
long long l = 0, r = 3e9, ans;
bool check(long long mid, long long need) {
  long long c = 2 * mid * mid + 2 * mid + 1;
  if (mid - x > 0) c -= (mid - x) * (mid - x);
  if (mid - y > 0) c -= (mid - y) * (mid - y);
  if (mid - (n - 1 - x) > 0) c -= (mid - (n - 1 - x)) * (mid - (n - 1 - x));
  if (mid - (n - 1 - y) > 0) c -= (mid - (n - 1 - y)) * (mid - (n - 1 - y));
  if (mid - x - (n - y - 1) - 1 > 0)
    c += (1 + mid - x - (n - y - 1) - 1) * (mid - x - (n - y - 1) - 1) / 2;
  if (mid - (n - 1 - x) - (n - 1 - y) > 0)
    c += (1 + mid - (n - 1 - x) - (n - 1 - y) - 1) *
         (mid - (n - 1 - x) - (n - 1 - y) - 1) / 2;
  if (mid - (n - 1 - x) - y - 1 > 0)
    c += (1 + mid - (n - 1 - x) - y - 1) * (mid - (n - 1 - x) - y - 1) / 2;
  if (mid - y - x - 1 > 0) c += (1 + mid - y - x - 1) * (mid - y - x - 1) / 2;
  return c >= need;
}
int main() {
  scanf("%lld%lld%lld%lld", &n, &x, &y, &c);
  x--;
  y--;
  while (l <= r) {
    long long mid = (l + r) >> 1;
    if (check(mid, c)) {
      ans = mid;
      r = mid - 1;
    } else {
      l = mid + 1;
    }
  }
  printf("%lld", ans);
  return 0;
}

```