#include <bits/stdc++.h>
using namespace std;
long long n, x, y, c;
inline long long f(long long x) { return x * x + 2 * x + 1; }
inline long long g(long long x) { return (x + 1) * (x + 2) / 2; }
long long solve(long long t) {
  long long ans;
  ans = 2 * t * t + 2 * t + 1;
  if (t >= x) ans -= f(t - x);
  if (t >= n + 1 - y) ans -= f(t - (n + 1 - y));
  if (t >= n + 1 - x) ans -= f(t - (n + 1 - x));
  if (t >= y) ans -= f(t - y);
  if (t >= n + 1 + x - y) ans += g(t - (n + 1 + x - y));
  if (t >= n + 1 - y + n + 1 - x) ans += g(t - (n + 1 - y + n + 1 - x));
  if (t >= n + 1 - x + y) ans += g(t - (n + 1 - x + y));
  if (t >= x + y) ans += g(t - (x + y));
  return ans;
}
int main() {
  scanf("%d%d%d%d", &n, &x, &y, &c);
  long long l = -1, r = 2 * n, m;
  while (l + 1 < r) {
    m = (l + r) / 2;
    if (solve(m) < c)
      l = m;
    else
      r = m;
  }
  cout << r;
  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, c;
inline long long f(long long x) { return x * x + 2 * x + 1; }
inline long long g(long long x) { return (x + 1) * (x + 2) / 2; }
long long solve(long long t) {
  long long ans;
  ans = 2 * t * t + 2 * t + 1;
  if (t >= x) ans -= f(t - x);
  if (t >= n + 1 - y) ans -= f(t - (n + 1 - y));
  if (t >= n + 1 - x) ans -= f(t - (n + 1 - x));
  if (t >= y) ans -= f(t - y);
  if (t >= n + 1 + x - y) ans += g(t - (n + 1 + x - y));
  if (t >= n + 1 - y + n + 1 - x) ans += g(t - (n + 1 - y + n + 1 - x));
  if (t >= n + 1 - x + y) ans += g(t - (n + 1 - x + y));
  if (t >= x + y) ans += g(t - (x + y));
  return ans;
}
int main() {
  scanf("%d%d%d%d", &n, &x, &y, &c);
  long long l = -1, r = 2 * n, m;
  while (l + 1 < r) {
    m = (l + r) / 2;
    if (solve(m) < c)
      l = m;
    else
      r = m;
  }
  cout << r;
  return 0;
}

```