#include <bits/stdc++.h>
using namespace std;
long long n, x, y, m;
long long cul(long long p) {
  if (p <= 0) return 0;
  return (p + 1) * p / 2;
}
int judge(long long p) {
  long long a = x - 1, b = y - 1, c = n - x, d = n - y;
  long long ans = min(p, a) + min(p, b) + min(p, c) + min(p, d) + 1;
  p--;
  ans += cul(p) * 4 - 2 * cul(p - a) - 2 * cul(p - b) - 2 * cul(p - c) -
         2 * cul(p - d);
  ans += cul(p - a - b) + cul(p - b - c) + cul(p - c - d) + cul(p - d - a);
  return ans >= m;
}
int main() {
  scanf("%lld%lld%lld%lld", &n, &x, &y, &m);
  long long l = 0, r = 1e9, ans = 0;
  while (l <= r) {
    long long mid = (r + l) >> 1;
    if (judge(mid)) {
      ans = mid;
      r = mid - 1;
    } else
      l = mid + 1;
  }
  printf("%lld\n", ans);
  return 0;
}

### Prompt

Develop a solution in CPP 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;
long long n, x, y, m;
long long cul(long long p) {
  if (p <= 0) return 0;
  return (p + 1) * p / 2;
}
int judge(long long p) {
  long long a = x - 1, b = y - 1, c = n - x, d = n - y;
  long long ans = min(p, a) + min(p, b) + min(p, c) + min(p, d) + 1;
  p--;
  ans += cul(p) * 4 - 2 * cul(p - a) - 2 * cul(p - b) - 2 * cul(p - c) -
         2 * cul(p - d);
  ans += cul(p - a - b) + cul(p - b - c) + cul(p - c - d) + cul(p - d - a);
  return ans >= m;
}
int main() {
  scanf("%lld%lld%lld%lld", &n, &x, &y, &m);
  long long l = 0, r = 1e9, ans = 0;
  while (l <= r) {
    long long mid = (r + l) >> 1;
    if (judge(mid)) {
      ans = mid;
      r = mid - 1;
    } else
      l = mid + 1;
  }
  printf("%lld\n", ans);
  return 0;
}

```