#include <bits/stdc++.h>
using namespace std;
long long n, x, y, c;
long long ca(long long x, long long y, long long mid) {
  long long t = mid - x + 1;
  if (t < 0) t = 0;
  long long ans = -t * t;
  t = mid - n - x + y;
  if (t < 0) t = 0;
  ans = ans + (1 + t) * t / 2;
  return ans;
}
long long calc(long long mid) {
  long long ans = 1 + mid * (1 + mid) * 2;
  ans = ans + ca(x, y, mid);
  ans = ans + ca(y, n + 1 - x, mid);
  ans = ans + ca(n + 1 - y, x, mid);
  ans = ans + ca(n + 1 - x, n + 1 - y, mid);
  return ans;
}
bool check(long long mid) { return calc(mid) >= c; }
int main() {
  while (~scanf("%I64d%I64d%I64d%I64d", &n, &x, &y, &c)) {
    long long l = 0, r = 3e9, ans = 0;
    while (l <= r) {
      long long mid = (l + r) >> 1;
      if (check(mid)) {
        ans = mid;
        r = mid - 1;
      } else {
        l = mid + 1;
      }
    }
    printf("%I64d\n", ans);
  }
  return 0;
}

### Prompt

Please formulate 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;
long long ca(long long x, long long y, long long mid) {
  long long t = mid - x + 1;
  if (t < 0) t = 0;
  long long ans = -t * t;
  t = mid - n - x + y;
  if (t < 0) t = 0;
  ans = ans + (1 + t) * t / 2;
  return ans;
}
long long calc(long long mid) {
  long long ans = 1 + mid * (1 + mid) * 2;
  ans = ans + ca(x, y, mid);
  ans = ans + ca(y, n + 1 - x, mid);
  ans = ans + ca(n + 1 - y, x, mid);
  ans = ans + ca(n + 1 - x, n + 1 - y, mid);
  return ans;
}
bool check(long long mid) { return calc(mid) >= c; }
int main() {
  while (~scanf("%I64d%I64d%I64d%I64d", &n, &x, &y, &c)) {
    long long l = 0, r = 3e9, ans = 0;
    while (l <= r) {
      long long mid = (l + r) >> 1;
      if (check(mid)) {
        ans = mid;
        r = mid - 1;
      } else {
        l = mid + 1;
      }
    }
    printf("%I64d\n", ans);
  }
  return 0;
}

```