#include <bits/stdc++.h>
using namespace std;
long long n, x, y, c, ans, switchedOn;
int i;
inline int next(int x) {
  x = (x + 1) % 4;
  return x;
}
int main() {
  scanf("%lld %lld %lld %lld", &n, &x, &y, &c);
  long long S[4] = {y - 1, x - 1, n - y, n - x};
  ans = 0, switchedOn = 1;
  while (1) {
    for (i = 0; i < 4; ++i) {
      switchedOn += ans;
      switchedOn -= 1LL * min(max(1LL * 0, ans - S[i]) +
                                  max(1LL * 0, ans - S[next(i)] - 1),
                              ans);
    }
    if (switchedOn >= c) break;
    ++ans;
  }
  printf("%lld\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;
long long n, x, y, c, ans, switchedOn;
int i;
inline int next(int x) {
  x = (x + 1) % 4;
  return x;
}
int main() {
  scanf("%lld %lld %lld %lld", &n, &x, &y, &c);
  long long S[4] = {y - 1, x - 1, n - y, n - x};
  ans = 0, switchedOn = 1;
  while (1) {
    for (i = 0; i < 4; ++i) {
      switchedOn += ans;
      switchedOn -= 1LL * min(max(1LL * 0, ans - S[i]) +
                                  max(1LL * 0, ans - S[next(i)] - 1),
                              ans);
    }
    if (switchedOn >= c) break;
    ++ans;
  }
  printf("%lld\n", ans);
}

```