#include <bits/stdc++.h>
using namespace std;
int n, x, y, c;
long long calc(long long n, long long m, long long k) {
  if (k <= 0) return 0;
  if (n <= 0 || m <= 0) return 0;
  if (n > m) swap(n, m);
  if (k <= n) return k * (k + 1) / 2;
  return min(k - n, m) * n + calc(n, m + n - k, n);
}
bool check(int t) {
  return calc(x, y, t + 1) + calc(n - x, n - y, t - 1) + calc(n - x, y, t) +
             calc(x, n - y, t) >=
         c;
}
int main() {
  scanf("%d%d%d%d", &n, &x, &y, &c);
  int l = -1, r = n + n + 1;
  int ans = -1;
  while (l <= r) {
    int mid = (l + r) >> 1;
    if (check(mid)) {
      ans = mid;
      r = mid - 1;
    } else
      l = mid + 1;
  }
  printf("%d\n", ans);
}

### Prompt

Your task is to create a Cpp solution to 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;
int n, x, y, c;
long long calc(long long n, long long m, long long k) {
  if (k <= 0) return 0;
  if (n <= 0 || m <= 0) return 0;
  if (n > m) swap(n, m);
  if (k <= n) return k * (k + 1) / 2;
  return min(k - n, m) * n + calc(n, m + n - k, n);
}
bool check(int t) {
  return calc(x, y, t + 1) + calc(n - x, n - y, t - 1) + calc(n - x, y, t) +
             calc(x, n - y, t) >=
         c;
}
int main() {
  scanf("%d%d%d%d", &n, &x, &y, &c);
  int l = -1, r = n + n + 1;
  int ans = -1;
  while (l <= r) {
    int mid = (l + r) >> 1;
    if (check(mid)) {
      ans = mid;
      r = mid - 1;
    } else
      l = mid + 1;
  }
  printf("%d\n", ans);
}

```